Question

Probability of Compound Events

Answer 1

Alrighty, so for the first question, we need to see how many different types of desserts are needed to result in 12 different combinations.

Answer 2

A customer can choose one of two sandwhiches, and one of two soups.

Using ONLY the sandwhiches and soups, we see that we get 4 different combinations (2 * 2 = 4)

Answer 3

So, how many desserts are available to make 12 different choices when all is said and done?

Answer 4

3

Answer 5

There ya go :)

Answer 6

on to the next one

Answer 7

For the next question, let's first find the probability for a single die to NOT roll to a 6. That is, what is the probability of it landing on a 1, 2, 3, 4, or 5?

Answer 8

1 over 6

Answer 9

That's the probability to land on a single number. We can use that, though.

There is a 1/6 probability of a single die landing on a 6.

We know:

\[P(Not6) = 1-P(6)\]
So what is the probability to land on anything other than a 6?

Answer 10

1 over 5

Answer 11

\[1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}\]

Answer 12

oh

Answer 13

Ok, so a single die has a 5 in 6 chance to NOT show a 6 when it's rolled.

So, now we want to find the probability that two dice will BOTH not show a 6.

We know:

\[P(AandB) = P(A)P(B)\]
For independent events. So what is the probability that BOTH of the dice will NOT show a 6?

Answer 14

10/12

Answer 15

\[\frac{5}{6} * \frac{5}{6} = ?\]

Answer 16

25 over 36

Answer 17

Ok, so there is a probability of 25/36 of neither dice showing a 6.

So the probability of AT LEAST one 6 showing is:

\[1-\frac{25}{36} = ?\]

Answer 18

11/36

Answer 19

There you go :)

To summarize, we found the probability for a single die to fail to show a 6. Using this, we found the probability for 2 die to fail to show a 6. So 1 minus the probability of both not showing a 6 gives the probability to show at least one six.

Make sense?

Answer 20

yup

Answer 21

For the next part, there are a few ways to do it. Are you familiar with binomial probability?

Answer 22

no, not really

Answer 23

Permutations and Combinations?

Answer 24

nope

Answer 25

Hmm, well, I guess the best way would just be to write out the different combinations of children, then. We'll assume that the chance for a child to be a girl is the same as the chance of a child to be a boy. If you list them from first born to last born, you can list the possible combinations, such as:

g,g,g,g
g,g,g,b
g,g,b,g
g,b,g,g
b,g,g,g
g,g,b,b

and so on.

You can then count the number of combinations that has 2 girls and 2 boys, and divide that by the total number of possible combinations.

Answer 26

If you know (or should know) how to use binomial probabilities, it's a lot easier.

\[\left(\begin{matrix}4 \\ 2\end{matrix}\right)[P(girl)]^2 [P(boy)]^2\]

Answer 27

well you could explain it ill get it

Answer 28

I can try.

The thing in front is called a "binomial coefficient." When you read that one, you'd read it as "4 choose 2."

What this does is lock in exactly how many girls (or boys) we want out of the group of 4, while not caring about what order they occur in (we don't care if the oldest 2 are girls and the youngest 2 are boys, or any such order).

\[\left(\begin{matrix}4 \\ 2\end{matrix}\right) = \frac{4!}{2! (4-2)!}\]

Answer 29

Normally, when we use this to get a probability, we define what a "success" is, and what a "failure" is. You can think of it that way if you wish.

To get a binomial probability, we just do this:

\[\left(\begin{matrix}N \\ k\end{matrix}\right) P(Success)^k P(Failure)^{N-k}\]

What this does is give us the probability of k successes (no more, no less), out of N events.

So, out of 4 children we want 2 boys and 2 girls. Call whichever you wish to be a "success," and the other a "failure," and we get:

\[\left(\begin{matrix}4 \\ 2\end{matrix}\right) P(Girl)^2 P(Boy)^2\]

Just plug in what you know, and the answer will appear after simplifying :)

Answer 30

so what do we multiply

Answer 31

Whats the probability of a child being a boy? Whats the probability of a child being a girl?

Answer 32

a boy is 2 over 4 and same with a girl

Answer 33

It's 1/2, correct. I think you're getting that from the wrong place, though. The size of the family has nothing to do with it. It's a boy, or a girl, and the probability is the same either way (50%).

So, P(Boy) = 1/2, and P(Girl) = 1/2

Plug that in:

\[\left(\begin{matrix}4 \\ 2\end{matrix}\right) (\frac{1}{2})^2 (\frac{1}{2})^{4-2}\]

Answer 34

Evaluate that, and you'll have your answer.

Answer 35

so first 1/2 times 1/2 times 1/2 times 1/2 is 1/16 and that times 4/2 is 1/8

Answer 36

\[\left(\begin{matrix}4 \\ 2\end{matrix}\right)\]
is your binomial coefficient. It's not 4/2

\[\left(\begin{matrix}4 \\ 2\end{matrix}\right) = \frac{4!}{2! (4-2)!} = \frac{4*3*2*1}{(2*1)(2*1)}\]

Answer 37

oh

Answer 38

how did you get 4 times 3

Answer 39

N! = N*(N-1)*(N-2)*...3*2*1

Answer 40

It's called a "factorial." It's denoted by an exclamation point after a number.

Answer 41

ok

Answer 42

can rewrite the whole binomial again

Answer 43

What do you mean?

Answer 44

explaining the binomial coefficent

Answer 45

so 24 over 2

Answer 46

Given N events, we want k successes, and we don't care what order they are in:

\[P(Successes = k) = \left(\begin{matrix}N \\ k\end{matrix}\right) P(Success)^k P(Failure)^{N-k}\]

\[\left(\begin{matrix}N \\ k\end{matrix}\right) = \frac{N!}{k!(N-k)!}\]

For this question, N = 4, k = 2, P(Success) = 1/2, P(Failure) = 1/2

Answer 47

\[\left(\begin{matrix}4 \\ 2\end{matrix}\right)(\frac{1}{2})^2 (\frac{1}{2})^2 = \frac{4*3*2*1}{2*1*2*1} \frac{1}{4} \frac{1}{4} = \frac{6}{16}\]

Answer 48

ohok

Answer 49

Next question, you flipped a coin twice. You are going to flip it again. Whats the probability that it will land heads?

Do the first two flips have any effect on flipping the coin again?

Answer 50

yes

Answer 51

How? And what is it?

Answer 52

1 over 2

Answer 53

Right. The first two flips do NOT have an effect on the third flip. The chance of heads on any individual flip is 1/2 (for a fair coin).

Answer 54

For the next question, I think they actually want you to list out the possible combinations. You'll just have to actually do that.

ggg
ggb
gbg
bgg

etc.

Answer 55

You can calculate the total number of choices fairly easily.

There are three children. Each one is either a boy or a girl (2 choices).

Total combinations: choices for first * choices for second * choices for third

\[2*2*2 = 2^3 = 8\]

Answer 56

so thats it

Answer 57

We still have the last question. I was going to let you write out the possible combinations before moving on.

Answer 58

ok i was talking about the combinations

Answer 59

Let me know when you have them. They make the next part easy.

Answer 60

ok

Answer 61

done

Answer 62

is the first one 1/8

Answer 63

Alright. Now, how many have "girl" in the first position, and either "boy" in the middle and "girl" in the last, or "girl" in the middle and "boy" in the last?

Answer 64

3

Answer 65

Not 3. Count carefully.

Your combinations should be:

ggg gbb
ggb bgb
gbg bbg
bgg bbb

Answer 66

4?

Answer 67

Nope. We want a "g" in the first position.

The other two are b and g in any order, but can't both be b, or both be g.

Answer 68

2

Answer 69

There ya go :)

So the probability is 2/8, or 1/4

Answer 70

ok for the next one is it 2 also

Answer 71

is the next one 3

Answer 72

Not 3. We want AT LEAST two girls. That is, there could be 2 or 3.

Answer 73

well thanks alot and for the last question i got a probability.

Answer 74

Yep. There are two combinations of boys and girls that only occur once (disregarding their position in the "lineup"). What are they?