MATH 213: SOLUTIONS FOR MIDTERM 2
1(10 points):A biased coin (the probability of getting Heads equals 1/3) is tossed
twice. Consider two random variables: let Xbe the number of Heads obtained, and
let Ybe equal to 1 if the two tosses yield the same result (both Heads or both Tails),
equal to 0 otherwise. Are the random variables Xand Yindependent?
Xis a binomial random variable: p(X=k) = 2
kpk(1 −p)2−k, hence p(X= 0) =
p(X= 1) = 4/9, p(X= 2) = 1/9. Moreover, p(Y= 1) = p(HH) + p(T T ) =
(1/3)2+ (2/3)2= 5/9, and p(Y= 0) = 1 −p(Y= 1) = 4/9. Note now that if Y= 0,
then X≤1. Then 0 = p(Y= 0, X = 2) 6=p(Y= 0)p(X= 2) = 4/81, and the
random variables Xand Yare not independent.
2(10 points):The final exam contains 50 true/false questions, each worth 2 points,
and 25 multiple-choice questions, each worth 4 points. Linda answers a true/false
question correctly with the probability of 0.6. The probability she answers a multiple-
choice question correctly is 0.8. Find Linda’s expected score.
Denote Linda’s score by X. Let X1(X2) be the number of correct answers on
the true-false section (resp. multiple-choice section). Then X= 2X1+ 4X2, hence
E(X) = 2E(X1) + 4E(X2). Note that X1is the number of successes in a series of
n= 50 independent Bernoulli trials, if the probability of success p= 0.6. Thus,
X1is a binomial random variable, and E(X1) = np = 50 ·0.6 = 30. Similarly,
E(X2) = 25 ·0.8 = 20. Then E(X) = 2 ·30 + 4 ·20 = 140.
3(10 points):What is the probability that a poker hand contains cards of five
different kinds?
Recall that a card deck has 52 cards, of 13 different kinds (denominations). There
are 4 suits, each containing one card of each denomination. A poker hand consists
of 5 cards.
Our sample space Sis the set of all selections of 5 cards (our of 52). The total
number of poker hands (that is, the number of collections of 5 cards) equals 52
5.
This is the cardinality of our sample space S. The event Eoccurs if 5 different kinds
of cards (out of 13) are present. There are 13
5ways to select the kinds of cards. For
each of the 5 selected kinds, there are four ways to pick a suit. Thus, |E|= 4513
5,
and p(E) = |E|/|S|= 4513
5/52
5.
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