Answer:
its down here ⤵⤵
Step-by-step explanation:
Math 116 - Solutions to Take-Home Midterm
1. A certain card game is played with a standard deck of 52 cards. You are dealt 7 cards,
and a winning hand consists of 4 cards of the same rank AND 3 consecutive cards in
the same suit (including the possibility of King - Ace - 2, etc.). What is the probability
of being dealt a winning hand?
Solution. The total number of possible hands is
527. To create a winning hand, we must choose one rank (from 13) in which we will have 4-of-a-kind. There are 13 ways to make this choice. Next, we must choose three consecutive ranks from the remaining 12 ranks. There are only 10 ways to do this, since we have to avoid the rank that we already chose for the 4-of-a-kind. Finally, we have to choose one of 4 suits for the three cards in a row, which can be done in 4 different ways. By the multiplication principle, we get a total of 13 ∗ 10 ∗ 4 = 520 winning hands. Hence, the probability of being dealta winning hand is 520/527≈ .000004.
2. You have 10 (indistinguishable) apples and 10 (indistinguishable) oranges.
(a) How many ways are there to pass all 20 pieces of fruit out to 10 children?
(b) How many ways are there to pass all 20 pieces of fruit out to 10 children so that each child gets 2 pieces?
Solution. a) The number of ways to pass out 10 apples to the 10 children is given by the number of 10-combinations with repetition of the 10-children, which is 10 + 10 − 1 10 − 1
=
19
9
. This is also the number of ways to pass out 10 oranges. Since any combination
of a distribution of the apples with a distribution of the oranges is possible, by the
multiplication principle, the total number of ways to distribute both the apples and
oranges (with no restrictions) is
19
9
2
.
b) Each child gets 2 pieces of fruit, and thus either 2 apples, 2 oranges, or 1 of each. We
will divide the problem into cases based on the number of children who get 1 of each.
Suppose that exactly k children get 1 orange and 1 apple. Obviously, 0 ≤ k ≤ 10. Also
notice that k must be even, since the remaining 10 − k apples must be handed out in
pairs. Now for any of these possible values of k, there will be
10
k
ways to choose
which k children get 1 of each type of fruit. Then there will be 10-k children left over,
