Question

Hello, some probability/combinations questions here, hopefully not too difficult but it seems I'm missing something important.

1. Suppose both sexes are equally probable. What is the probability that a family of 4 children has exactly 3 boys?

2. 5 cards are dealt from a shuffled deck of 52 cards. What is the probability that a hand contains all hearts in a sequence from an ace to 10?

3. A Motor Vehicles Branch has 5 different written driving tests to administer randomly to prospective drivers. 2 women and 3 men take the test. What is the probability that exactly 3 people take the same

Answer 1

Text got cut off, #3 asks for the probability that exactly 3 people take the same test. For the first question I thought it would be (1/2)^4 = 1/16, but the answer is 1/4. The second and third I'm not sure where to begin other than that the denominator in #2 will be 52C5.

Answer 2

For #1, it's just \[\left(\frac{1}{2}\right)^4 \binom{4}{1}=\frac{4}{16}=\frac{1}{4}\]You multiply by \(\binom41\) because the single girl in the family could have been born 1st, 2nd, 3rd, or 4th.

Answer 3

I see, thanks. I also just figured out it can be done using binomial theorem (although your answer looks more simple): 4C3 * (1/2)^3 * (1/2)^1 ... they're the same thing really since 4C3=4C1.

Answer 4

As for the second one, my understanding is that it has 5 cards, all of which are in the set {ace, 2, 3, 4, 5, 6, 7, 8, 9, 10} of hearts, and they have to be in a sequence (i.e., 2,3,4,5,6 or 5,6,7,8,9). Is this this also what you think?

Answer 5

Yes, I believe that's it. The question perhaps isn't worded the best way but it doesn't make sense for it to be all cards in the set since you can only have 5 cards.

Answer 6

In that case, how many sequences containing only 5 numbers can be made of the set {1,2,3,4,5,6,7,8,9,10}? Where the sequence is of the form \(a_n+1\).

Answer 7

it is prob ...A,K,Q,J,10...from ace to 10 (5 total cards)

Answer 8

I was wondering if that was the case.

Answer 9

@KingGeorge I'm not sure what you mean by "an+1" but I believe it would be 10C5?

Answer 10

Wait, maybe not, that would be to choose 5 individual numbers..

Answer 11

What I mean is that {1,2,3,4,5} and {2,3,4,5,6} are possible, but {1,3,4,5,6} is not.

Answer 12

Ah, right, then just by counting it looks like there would be 6 possibilities: {1,2,3,4,5}, {2,3,4,5,6}, {3,4,5,6,7}, {4,5,6,7,8}, {5,6,7,8,9}, and {6,7,8,9,10}

Answer 13

Right, and if we instead use the set {10, J, Q, K, ace} how many are there?

Answer 14

7, right? But I just tried 6/(52C5) and 7/(52C5), none of which give the answer..

Answer 15

If we were to use the set {10, J, Q, K, ace} there's only one possibility of choices. Namely, itself {10,J,Q,K,A}.

Answer 16

Ah, that's it, 1/(52C5). And I assume that's because the question is actually asking for the set {10,J,Q,K,A}, not the others?

Thanks anyway, I'd give 10 medals if I could ^^

Answer 17

That's exactly why it's 1 and not 6.