The probability to get a head once is $\frac{1}{2}$ so the probability that she gets it three times in a row is $\frac{1}{2} \cdot \frac{1}{ 2} \cdot \frac{1}{2} = \pmb{\frac{1}{8}}$

The probability that she scores the ball is $\frac{3}{5}$ and the probability that she misses is $\frac{2}{5}$. So the probability that she scores, misses and then scores is $\frac{3}{5} \cdot \frac{2}{5} \cdot \frac{3}{5} = \pmb{\frac{18}{125}}$.

The probability that the first marble is green is $\frac{6}{10}$. The second time Jacob picks a marble there are only $9$ left. So the probability that the second marble is green is $\frac{5}{9}$. The probability that Jacob picks two green marbles is therefore $\frac{6}{10} \cdot \frac{5}{9} = \frac{30}{90} = \pmb{\frac{1}{3}}$.

a) It’s impossible for it to rain unless it is cloudy. The events are dependent.

b) The outcome of one die roll doesn’t affect the probability of getting an even number on a second die roll. The events are independent.

c) Making a left and a right turn cannot occur at the same time. The events are dependent.

d) The probability of going to jail is obviously much higher if we rob a bank compared to if we do not. The events are dependent.

e) The events of getting stuck in traffic and winning the lottery obviously have no impact on one another. The events are independent.

Two events are independent if and only if $P(A) = P(B \ | \ A)$ or $P(B) = P(B \ | \ A)$. The general multiplication rule states that:

$$P(A) \ \cdot \ P(B \ | \ A) \ = \ P(A \ \text{and} \ B)$$

If we compare this with the equation given in the question, we can see that $P(B) = P(B \ | \ A)$. So if: 

$$P(A) \ \cdot \ P(B) \ = \ P(A \ \text{and} \ B)$$ 

we can be $100$% certain that the events $A$ and $B$ are independent. 

The events of sunny and windy weather are independent if and only if $P(\text{sunny}) \cdot P(\text{windy}) =$ $P(\text{sunny and windy})$. But from the question we know that $P(\text{sunny}) \cdot P(\text{windy}) =$ $\frac{2}{5} \cdot \frac{5}{12} = \frac{1}{6}$ while $P(\text{sunny and windy}) \ =$ $\frac{1}{7}$. The events of sunny and windy weather are therefore dependent.

The probability that the first student is a girl is $\frac{8}{17}$. Next time we select a student there are only $16$ children left, so the chance of that student being a boy is $\frac{9}{16}$. When we choose the third student there are $8$ boys left from a total of $15$ students, so the probability that we choose another boy is $\frac{8}{15}$. It therefore follows that $P(\text{girl-boy-boy}) = $ $\frac{8}{17} \cdot \frac{9}{16} \cdot \frac{8}{15} \ =$ $\frac{1}{17} \cdot \frac{3}{1} \cdot \frac{4}{5} \ =$ $\pmb{\frac{12}{85}}$.

$$ P(A) = P(\text{only} \ A) + P(A \ \text{and} \ B) \\ P(A) = \frac{15}{60} + \frac{9}{60} = \pmb{\frac{2}{5}}$$

$$ P(B) = P(\text{only} \ B) + P(A \ \text{and} \ B) \\ P(B) = \frac{1}{60} + \frac{9}{60} = \pmb{\frac{1}{6}}$$

We can use the general multiplication rule to help us solve this question: 

$$P(A) \ \cdot \ P(B \ | \ A) \ = \ P(A \ \text{and} \ B) \\ P(B) \ \cdot \ P(A \ | \ B) \ = \ P(A \ \text{and} \ B)$$

Using the given probabilities we get:

$$\Big( \dfrac{15}{60} + \dfrac{9}{60} \Big) \ \cdot \ P(B \ | \ A) \ = \ \dfrac{9}{60}$$

$$P(B \ | \ A) \ = \ \dfrac{\Big(\dfrac{9}{60}\Big)}{\Big(\dfrac{24}{60}\Big)}$$

$$P(B \ | \ A) \ = \ \pmb{\dfrac{3}{8}}$$

And

$$\Big( \dfrac{1}{60} + \dfrac{9}{60} \Big) \ \cdot \ P(A \ | \ B) \ = \ \dfrac{9}{60}$$

$$P(A \ | \ B) \ = \ \dfrac{\Big(\dfrac{9}{60}\Big)}{\Big(\dfrac{10}{60}\Big)}$$

$$P(A \ | \ B) \ = \ \pmb{\dfrac{9}{10}}$$

Two events $A$ and $B$ are independent if and only if:

$$P(A) \ \cdot P(A) \ = \ P(A \ \text{and} \ B)$$

In this case we have: 

$$P(A) \ = \ \Big( \dfrac{15}{60} + \dfrac{9}{60} \Big) \ = \ \dfrac{3}{8}$$

$$P(B) \ = \ \Big( \dfrac{1}{60} + \dfrac{9}{60} \Big) \ = \ \dfrac{1}{6}$$

$$P(A \ \text{and} \ B) = \frac{9}{60} \ = \ \dfrac{3}{20}$$

And since:

$$\dfrac{3}{8} \ \cdot \ \dfrac{1}{6} \ = \ \dfrac{1}{16} \ \neq \ \frac{3}{20}$$

the events are dependent. 

All fractions above have $60$ as a common denominator. We are therefore going to express all probabilities with denominator $60$ to make our calculations easier to follow:

$P(\text{only} \ A) \ = \ \frac{1}{12} \ = \ \frac{5}{60}$ 

$P(A, B \ \text{and} \ C) \ = \ \frac{1}{15} \ = \ \frac{4}{60}$

$P(\text{neither} \ A, B \ \text{or} \ C) = \frac{1}{6} \ = \frac{10}{60}$ 

All the three regions above can be filled in immediately as none of them overlap with any other regions. 

From the information given in the question we know that $P(A \ \text{and} \ C) = \frac{13}{60}$. Since $P(A, B \ \text{and} \ C) = \frac{4}{60}$ it follows that: 

$$P(A \ \text{and} \ C \ \text{but not} \ B) = \frac{13}{60} - \frac{4}{60} = \pmb{\frac{9}{60}}$$

Let $x = P(A \ \text{and} \ B \ \text{but not} \ C)$. We can then fill in the venn diagram as follows:

It now follows that:

$$ P(A) = \frac{5}{60} + \frac{9}{60} + \frac{4}{60} + x = \frac{18}{60} + x $$

But since we know that $P(A) = \frac{7}{15} = \frac{28}{60}$ we have:

$$\frac{18}{60} + x \ = \ \frac{28}{60}$$

$$ x \ = \ \pmb{\frac{10}{60}}$$

Let $y = P(B \ \text{and} \ C \ \text{but not} \ A)$. We can now fill in the venn diagram as follows:

If we now add the circles $B$ and $C$, subtract the region where both circles overlap (because we counted it twice) and then add all remaining regions outside of circles $B$ and $C$ we added every region in the entire diagram. Because the diagram represents every possible outcome that can occur, the sum of all these region must equal $1$ or $\frac{60}{60}$. This gives us:

$$ P(B) + P(C) - \Big( \frac{4}{60} + y \Big) + \frac{5}{60} + \frac{10}{60} \ = \ \frac{60}{60}$$

$$ P(B) + P(C) - y \ = \ \frac{49}{60}$$

We know that $P(B) = \frac{23}{60}$ and that $P(C) = \frac{7}{15} = \frac{28}{60}$. If we input this into the equation we get:

$$\frac{23}{60} + \frac{28}{60} - y \ = \ \frac{49}{60}$$

$$ y = \frac{2}{60}$$

It then follows that:

$$ P(B) \ = \ \frac{23}{60} - \frac{10}{60} - \frac{4}{60} - \frac{2}{60} \ = \ \frac{7}{60}$$

$$ P(C) \ = \ \frac{28}{60} - \frac{9}{60} - \frac{4}{ 60} - \frac{2}{60} \ = \ \frac{13}{60}$$

So the complete venn diagram looks like this:

We can solve this by determining $P(\text{green-green})$, $P(\text{green-blue})$ and $P(\text{blue-green})$ and then add the probability of each outcome. But we can also determine the probability that it will not happen, $P(\text{blue-blue})$, and then subtract this from $1$. Not choosing two blue marbles is the same as choosing at least one green marble. The probability that we choose two blue marbles is $\frac{2}{15}$, so the probability that we choose at least one green marble is therefore $1 - \frac{2}{15} = \pmb{\frac{13}{15}}$.

The probability that Sandra gets no red lights is $\dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} =$$\pmb{\dfrac{1}{64}}$. 

Case (3), that Sandra gets exactly $1$ red light, is a little more tricky. If you look at the tree diagram you’ll see that there are three ways in which Sandra could get exactly $1$ red light: 

(green-green-red) $=\dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{3}{4} = \dfrac{3}{64}$

(green-red-green) $=\dfrac{1}{4} \cdot \dfrac{3}{4} \cdot \dfrac{1}{4} = \dfrac{3}{64}$

(red-green-green) $=\dfrac{3}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} = \dfrac{3}{64}$

The probability that she gets exactly $1$ red light is the sum of all these three sub-cases. Notice that the likelihood of each case is the same since we are multiplying the same factors, only in a different order. This means that we can just multiply the by $3$ instead of adding all sub-cases one by one. 

The probability that Sandra gets exactly $1$ red light is therefore $3 \ \cdot $$\Big(\dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{3}{4} \Big) =$$ \pmb{\dfrac{9}{64}}$. 

How many ways can we arrange $1$ red and $2$ green traffic lights? The answer is $\dfrac{3!}{2!}=\dfrac{3 \cdot 2 \cdot 1}{2 \cdot 1}=3$ ways! Three unique objects can be arranged in $3\cdot2\cdot1} ways, and since two of them are identical in this case we need to divide by $2 \cdot 1$ because it doesn't matter in which order they come in. 

When you need to determine an outcome that can occur in several different ways, you can always use your knowledge to make your calculations faster. If you are not familiar with these methods then check out the lesson on lesson on combinatorics. 

The probability of getting either an even or odd number is $\frac{1}{2}$. This means that $P(\text{even-even-even-odd-odd}) = $$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} =$$ \Big( \frac{1}{2} \Big)^5 =$$ \frac{1}{64}$. However, this outcome can occur in $\frac{5!}{3! \ \cdot \ 2!} = 10$ different ways. The answer is therefore $\frac{1}{64} \cdot 10 = \pmb{\frac{5}{32}}$.

The probability of getting two heads and one tails is $\big(\frac{1}{2} \Big)^3 \cdot \frac{3!}{2!} =$$ \frac{3}{8}$ and the probability of getting three heads is $\big(\frac{1}{2} \Big)^3 = $$\frac{1}{8}$. Patrick's statement is true.

Now let’s turn to Tobias' statement. Coin flips are independent events, so the probability of getting tails on the third flip is still $\frac{1}{2}$. Before we began flipping the coin, there were three ways in which we could get two heads and one tails: 

(heads-heads-tails)

(heads-tails-heads)

(tails-heads-heads). 

However, when we have flipped the coin twice and gotten two heads, there is only one way to get the desired outcome: (heads-heads-tails). This is the difference between Patrick's and Tobias' reasoning. 

If you flip a coin and get heads again and again, it’s common to think "I’ve already gotten so many heads. The next flip must be tails!". But in fact, it is just as likely to get heads as tails on the next flip no matter how many heads you already have. When events are independent, past outcomes have no effect on future outcomes. Tobias' statement is false. 

There are three possible outcomes where the first marble is white: 

(white marble #$1$ from the bag with two white marbles) 

(white marble #$2$ from the bag with two white marbles)

(white marble from the bag with one white and one black marble)

It is equally likely that any of these three cases will occur. If either of the first two cases happens, the second marble will also be white. The probability that we choose two white balls is therefore $\pmb{\frac{2}{3}}$.

We make two decisions in this problem. First, we decide which of the three doors we want to choose. Second, we decide whether we want to keep our initial choice or not. There are therefore six possible outcomes in this problem:

(Goat #$ 1 $ - keep)    $\Longrightarrow $   Goat

(Goat #$ 1 $ - switch)    $\Longrightarrow $   Ferrari

(Goat #$ 2 $ - keep)    $\Longrightarrow $   Goat

(Goat #$ 2 $ - switch)    $\Longrightarrow $   Ferrari

(Ferrari - keep)    $\Longrightarrow $   Ferrari

(Ferrari - switch)     $\Longrightarrow $   Goat

If we decide to keep our initial choice we will get a goat in two of the three possible cases $\Big( \dfrac{1}{3} \Big)$. But if we switch doors we get a Ferrari in two out of three cases $\Big( \dfrac{2}{3} \Big)$. The best decision is therefore to switch doors.

Another way to think about this is that we have a $\frac{2}{3}$ chance of picking a goat in the first step. So if we decide to keep our choice the probability is $\frac{2}{3}$ that we get a goat and $\frac{1}{3}$ that we get the Ferrari. Look at the different outcomes outlined above so you really understand this. If we switch doors, however, we turn these probabilities around so that the likelihood of getting the Ferrari is $\frac{2}{3}$ while the probability of getting a goat is $\frac{1}{3}$.

For the sum of the two numbers to be even, both must be either odd or even. We have:

$P(\text{both even} = \frac{1}{4} \cdot \frac{2}{3} =$$ \frac{1}{6}$ 

$P(\text{both odd} =$\frac{3}{4} \cdot \frac{1}{4} =$$ \frac{1}{4}$

So the probability that the sum is even is $\frac{1}{4} \cdot \frac{1}{6} = $$\pmb{\frac{5}{12}}$.

We have three possible outcomes in this problem:

(1) $\pmb{0}$ red cards

(2) $\pmb{1}$ red card

(3) $\pmb{2}$ red cards 

The probability of getting a red card is the sum of (2) and (3). In this situation it’s simplest to use the subtraction method, so we will determine the probability of not getting a red card. 

The odds of getting $0$ red cards are $\frac{2}{4} \cdot \frac{1}{3} = \frac{1}{6}$. The probability of getting a red card is therefore $ 1 - \frac{1}{6} = \pmb{\frac{5}{6}}$.

At any given moment there are either:

(1) $\pmb{0}$ bird mothers

(2) $\pmb{1}$ bird mother

(3) $\pmb{2}$ bird mothers

(4) $\pmb{3}$ bird mothers

(5) $\pmb{4}$ bird mothers 

in the forest clearing.

The probability that there is at least one bird mother the forest clearing is the sum of cases (2), (3), (4) and (5). In this situation it’s obviously easier to use the subtraction method. We will therefore determine case (1) and subtract this from $1$. 

A bird mother spends $48$ minutes every hour away from the nest. So the likelihood of a specific bird mother not being in the forest clearing is $\frac{48}{60} = \frac{4}{5}$. The probability that none of the four bird mothers are present is $ \Big(\frac{4}{5} \Big)^5 = \frac{256}{625}$. 

The probability that at least one bird mother is in the clearing at a given moment is therefore $ 1 - \frac{256}{625} = \pmb{\frac{369}{625}}$.

he first ant can choose to go either clockwise or counter clockwise. The other four ants must all choose to go in the same direction as the first ant to avoid colliding. Each of these four ants has two directions to choose from so the probability that they never collide is $ \Big(\frac{1}{2} \Big)^4 = \pmb{\frac{1}{16}}$.

If we toss a coin three times we can get $ 2^3 = 8 $ possible outcomes. In three of these outcomes (heads-tails-tails, tails-tails-heads, tails-tails-tails) we get at least two tails in a row. The probability of getting at least two tails in a row is thus $ \pmb{\frac{3}{8}}$.

The difference between two consecutive balls can be either $ 1 $, $ 2 $, $ 3 $ or $ 4 $. That the difference is $ 1 $ can happen in seven ways, that the difference is $ 2 $ can happen in five ways, that the difference is $ 3 $ can happen in three ways and that the difference is $ 4 $ can happen in one way. In total, there are $ 7 + 5 + 3 + 1 = 16 $ ways that the differences can be equal. One way to illustrate why we end up with $7,5,3$ and $1$ ways for the differences to be equal is to draw nine empty spaces next to each other. How many ways you could place three objects in these empty spaces if the objects on the right and left both must be either one, two, three or four spaces away from the center object? 

We can choose three balls from an urn with nine balls in $\displaystyle{\binom{9}{3}}=$$\frac{9 \ \cdot \ 8 \ \cdot \ 7}{3!} = 84 $ different ways. The probability that the difference between consecutive balls is equal is thus $\frac{16}{84} = \pmb{\frac{4}{21}}$.

a) That exactly one flower sprouts can happen in five different ways: it’s either flower #$1$, #$2$, #$3$, #$4$ or #$5$. The probability that any particular one of these five outcomes will occur is $\frac{3}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} =$$ \frac{3}{1024}$. The probability that exactly one flower sprouts is therefore $ \pmb{\frac{15}{1024}}$.

b) That exactly two seeds sprout can happen in $\frac{5 \ \cdot \ 4}{2!} = 10 $ different ways. The probability that any particular of these ten outcomes will occur is $\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} =$$ \frac{9}{1024}$, so we have to multiply this by $ 10 $. The probability that exactly two of the five flowers sprout is thus $\frac{90}{1024} = \pmb{\frac{45}{512}}$.

c) We can divide this problem into the following outcomes:

(1) $\pmb{0}$ flowers sprouts

(2) $\pmb{1}$ flower sprout

(3) $\pmb{2}$ flowers sprouts

(4) $\pmb{3}$ flowers sprouts

(5) $\pmb{4}$ flowers sprouts

(6) $\pmb{5}$ flowers sprouts

The probability that at most three flowers sprout is the sum of cases (1), (2), (3) and (4). In this situation, it’s simpler to use the subtraction method and instead determine the sum of cases (5) and (6) and subtract the result from $1$. 

That exactly four flowers that sprout can happen in $\frac{5 \cdot 4 \cdot 3 \cdot 2}{4!} = 5$ different ways. The probability that any particular of these five outcomes occurs is $\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} = $$ \frac{81}{1024}$. The probability that exactly four flowers sprout is therefore $\frac{81}{1024} \cdot 5 = \frac{405}{1024}$.

That all flowers sprout can only happen in only one way. The probability of this outcome is therefore $ \Big(\frac{3}{4} \Big)^5 = \frac{243}{1024}$.

The sum of these two outcomes is $\frac{405}{1024} + \frac{243}{1024} = \frac{648}{1024}$. The probability that we get at most three flowers is thus $ 1 - \frac{405}{1024} - $$\frac{243}{1024} - =$$ \frac{376}{1024} =$$ \pmb{\frac{47}{128}}$.

$ 247 $ normal births and $ 3 $ twin births means $247 \cdot 1 + 3 \cdot 2 =$$ 253 $ children. $ 6 $ of these children are twins. This means that there are $\frac{6}{253} \cdot 9.6 \cdot 10^6 \ approx \pmb{230 \ 000}$ twins in Sweden.

a) When a number is divided by $ 3 $ we get one of three possible remainders: $ 0 $, $ 1 $ or $ 2 $. In order for the sum of the numbers to be divisible by $ 3 $, the sum of the remainders of those numbers when divided by $3$ must be divisible by $ 3 $. Once we have selected the first number, there is only one remainder which the second number may have for the sum to divisible by $ 3 $. So the answer is $ \pmb{\frac{1}{3}}$.

b) Since every third number is divisible by $ 3 $, the probability that we choose such a number is $\frac{1}{3}$. For the product of the numbers to be divisible by $ 3 $, at least one of the numbers must be divisible by $ 3 $. We can divide this problem into three possible outcomes:

(1) None of the numbers are divisible by $\pmb{3}$

(2) Exactly one of the numbers are divisible by $\pmb{3}$ 

(3) Both of the numbers are divisible by $\pmb{3}$

We want to find the sum of cases (2) and (3). The simplest approach here is to use the subtraction method and instead determine (1): the probability that none of the numbers are divisible by $3$. The answer is therefore:

$$ P(\text{at least one number divisible by} \ 3) \ = $$

$$ 1 - P(\text{no number divisible by} \ 3) \ = $$

$$ 1 - \frac{2}{3} \cdot \frac{2}{3} \ =$$ 

$$\pmb{\frac{5}{9}}$$

The frog sitting on the stump can be either a female or male frog. Since there are as many frogs of each sex, the probability is that that frog is a female is $\frac{1}{2}$. 

There are three possible cases for the frogs sitting in the clearing: (female-male), (male-female) or (male-male). Because each fall is equally likely to occur, the probability that one of the frogs is a female is $\frac{2}{3}$. So you should run left to maximize your chance of survival.

The baseball player can get two hits and two misses in $\frac{4!}{2! \ \cdot \ 2!} = 6 $ different ways and the probability of each of these outcomes is $\frac{3}{10} \cdot \frac{3}{10} \cdot \frac{7}{10 } \cdot \frac{7}{10} = $$\frac{441}{10000}$. The probability that she gets exactly two hits and two misses is therefore $ 6 \cdot \frac{441}{10000} = $$\pmb{\frac{1323}{5000}}$.

If we roll three dice we can get the following four outcomes:

(1) $\pmb{0}$ sixes

(2) $\pmb{1}$ six

(3) $\pmb{2}$ sixes

(4) $\pmb{3}$ sixes

The probability of getting at most one six is equal to the probabilities of cases (1) and (2). The probability of (1) is $ \Big(\frac{5}{6} \Big)^3 = \frac{125}{216}$ and the probability of (2) is $\frac{1}{6} \cdot \Big(\frac{5}{6} \Big)^2 \cdot \frac{3!}{2!} = \frac{75}{216}$. (We multiply with $\frac{3!}{2!}$ becasue that is how many ways that one six and two non-sixes can be arranged). We know have:

$$ P(\text{at most one six on three throws}) = $$

$$\frac{125}{216} + \frac{75}{216} = \frac{200}{216}$$ 

If we roll four dice, we can get the following five outcomes:

(1) $\pmb{0}$ sixes

(2) $\pmb{1}$ six

(3) $\pmb{2}$ sixes

(4) $\pmb{3}$ sixes

(5) $\pmb{4}$ sixes

The probability of getting at most two sixes is equal to $1 - P(3 \ \text{sixes}) - $$P(4 \ \text{sixes})$. 

$P(3 \ \text{sixes}) \ =$$ \Big(\frac{1}{6} \Big)^3 \cdot \frac{5}{6} \cdot \frac{4!}{3!} =$$ \frac { 20}{1296}$ 

$P(4 \ \text{sixes} \ =$$ \Big(\frac{1}{6} \Big)^4 = \frac{1}{1296}$. 

We then have:

$$ P(\text{at most two sixes on three throws}) = $$

$$1 - \Big(\frac{21}{1296} + \frac{1}{1296} \Big) = $$

$$\frac{1274}{1296}$$

In comparison, $ P(\text{at most one six on three throws}) = $$\frac{200}{216} = $$\frac{1200}{1296}$. It is therefore more likely to get at most two sixes on a four-dice roll than to get at most one six on a three-dice roll. Your friend's statement is false.

Since the probability of getting at least one even number on two dice rolls is $\frac{1}{3}$, the probability of getting two odd numbers must be $\frac{2}{3}$. Die rolls are independent events, so this means means that:

$$ P(\text{odd}) \cdot P(\text{odd}) = \frac{2}{3}$$

$$ P(\text{odd}) = \pmb{\sqrt{\frac{2}{3}}}$$

We can solve this problem by determining number of five-card shuffles that contain exactly two aces and dividing it by the total number of possible five-card shuffles.

There are $\displaystyle{\binom{4}{2}}=$$\frac{4 \ \cdot \ 3}{2!} = 6 $ ways to choose two of the four aces. There are $48$ other non-ace cards and we can choose three of these in $\displaystyle{\binom{48}{3}}=$$\frac{48 \ \cdot \ 47 \ \cdot \ 46}{3!} = $$(16 \cdot 47 \cdot 23) $ ways . The number of five-card shuffles with exactly two aces is thus $ 6 \cdot (16 \cdot 47 \cdot 23) $. We will wait with multiplying this out because we will also divide the result by the total number of five-card shuffles. We will likely be able to eliminate several common factors when we perform that division. It’s therefore better to wait with multiplying the numbers until we have been able to simplify the fraction as far as possible. 

The total number of possible five-card shuffles is:

$$displaystyle{\binom{52}{5}} = \dfrac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}{5!} = $$

$$\frac{52}{2} \cdot \frac{51}{ 3} \cdot \frac{50}{5} \cdot \frac{49}{1} \cdot \frac{48}{4} = $$

$$ 26 \cdot 17 \cdot 10 \cdot 49 \cdot 12 $$

Notice how we also simplified this fraction as far as possible before multiplying out the numerator and the denominator.

The probability that a five-card deck contains exactly two aces is therefore:

$$\frac{6 \cdot 16 \cdot 47 \cdot 23}{26 \cdot 17 \cdot 10 \cdot 49 \cdot 12} = $$

$$\frac{6 \cdot 47 \cdot 23}{13 \cdot 17 \cdot 5 \cdot 49 \cdot 3} = $$

$$\frac{2 \cdot 47 \cdot 23}{13 \cdot 17 \cdot 49} = $$

$$ \pmb{\frac{2162}{10829}}$$

Here you can see the benefit of keeping the numerator and denominator as a product of separate numbers rather than multiplying them the first thing we do. Had we multiplied the fraction to $\frac{103 \ 776}{2 \ 598 \ 960}$ instead of $\frac{6 \ \cdot \ 16 \cdot \ 47 \ \cdot \ 23}{26 \ \cdot \ 17 \ \cdot \ 10 \ \cdot \ 49 \ \cdot \ 12}$ it would have been much harder to see that both the numerator and the denominator were divisible by $48$. It’s almost always easiest to keep your results as a product until the end .

In this problem, it's ok to leave $\frac{6 \ \cdot \ 16 \cdot \ 47 \ \cdot \ 23}{26 \ \cdot \ 17 \ \cdot \ 10 \ \cdot \ 49 \ \cdot \ 12}$ as your final answer. The most important thing is that you show how you arrived at the answer and that you understand how to solve the problem. Not that you can multiply with large numbers. If the numbers become large, it’s perfectly ok to leave the answer as an incomplete calculation.

The probability that Linda both works at a bank and is active in the feminist movement can be written as $ P(\text{bank}) \cdot P(\text{feminist}) $. This is obviously less compared to the probability that she works at a bank, $ P(\text{bank})$. Option a) is thus most likely. 

We can illustrate this situation with a venn diagram. Let circle $ B $ be the probability that Linda works at a bank and let circle $ F $ be the probability that Linda is active in the feminist movement. Option a), the probability that Linda works at a bank, is then the whole circle $ B $.

Option b), the probability that Linda both works at a bank and is active in the feminist movement, is only the intersection of the circles $ B $ and $ F $.

Since b) is a subset of a), it is obviously more likely that a) is true than that b) is true. If b) is true, then by definition a) must be true as well. But a) may be true without b) being true.

Let us list the last digits for $ 11^x, 13^x, 15^x 17^x $ and $ 19x $ where $ x = 1, 2, 3, 4… $:

$ \text{Last digit} \ 11^x: \ \ \ \ \ 1, 1, 1, 1, 1… $

$ \text{Last digit} \ 13^x: \ \ \ \ \ 3, 9, 7, 1, 3… $

$ \text{Last digit} \ 15^x: \ \ \ \ \ 5, 5, 5, 5, 5… $

$ \text{Last digit} \ 17^x: \ \ \ \ \ 7, 9, 3, 1, 7… $

$ \text{Last digit} \ 19^x: \ \ \ \ \ 9, 1, 9, 1, 9… $

$ 11^x $ always ends in $ 1 $. $ 15^x $ never has $1$ as its last digit. $ 13^x $ and $ 17^x $ end in the digit $ 1 $ if and only if $ x $ is divisible by $ 4 $, and $ 19x $ ends with a $ 1 $ if and only if $ x $ is divisible by $ 2 $. Of the total $ 20 $ numbers that $ q $ is selected from, five of them are divisible by $ 4 $ and ten are divisible by $ 2 $. The probability that $ p^q $ ends with a $ 1 $ is thus:

$ \Big(\dfrac{1}{5} \cdot 1 \Big) + \Big(\dfrac{1}{5} \cdot \dfrac{5}{20} \Big) + $$\Big(\dfrac{1}{5} \cdot 0 \Big) + \Big(\dfrac{1}{5} \cdot \dfrac{10}{20} \Big) \ =$$ \ \pmb{\dfrac{2}{5}}$

We will solve this problem by calculating the number of ways that exactly six of the letters can reach the right person and then dividing it by the total number. possible ways in which the letters can be sent to the various persons.

If exactly six of the letters reach the right person, it means that exactly two of the letters do not. We can select these two people in $\frac{8 \ \cdot \ 7}{2!} = 28 $ ways. The total number of ways the letters can be sent is $ 8! $. The answer to the problem is therefore:

$$\frac{28}{8!} = $$

$$\frac{7 \ \cdot \ 4}{8 \ \cdot \ 7 \ \cdot \ 6 \ \cdot \ 5 \ \cdot \ 4 \ \cdot \ 3 \ \cdot \ 2 \ \cdot \ 1} = $$

$$\frac{1}{8 \ \cdot \ 6 \ \cdot \ 5 \ \cdot \ 3 \ \cdot \ 2 \ \cdot \ 1} = $$

$$ \pmb{\frac{1}{1440}}$$

Take a closer look at the fraction $\frac{609}{625}$. Why do we have these two specific numbers? $ 625 $ is $ 5 \cdot 5 \cdot 5 \cdot 5 = 5^4$. The difference $625 - 609$ also happens to be $16$, which is the same as $2 \cdot 2 \cdot 2 \cdot 2 = 2^4$. Try to use this information to solve the problem on your own before you continue to read the rest of the solution.

Let the probability of seeing a car in the next $ 5 $ minutes be $ p $. The odds of not seeing a car in the next $ 5 $ minutes are then $ (1-p). $ This means that the probability of not seeing a car in the next $ 20 $ minutes is $ (1-p)^4 $. We also know that the probability that Richard will not see a car in the next $ 20 $ minutes is $\frac{16}{625}$. This gives us:

$$ (1 - p)^4 = \frac{16}{625}$$

$$ (1 - p) = \frac{2}{5}$$

$$\frac{3}{5 } = p $$ 

The probability that Richard will see a car in the next $ 5 $ minutes is thus $ \pmb{\frac{3}{5}}$.