Question: a) How many diamonds would you expect to be drawn if 9 cards are randomly selected from the deck?
b) How many cards that are multiples of 3 cards would you expect to be drawn if 4 cards are randomly selected from the deck?
Can someone give some idea how to solve this? Not sure if this question means one after other or all 4 cards at the same time?
$\endgroup$ 02 Answers
$\begingroup$a) Let $X$ be the total number of diamonds drawn, and let $X_1,\ldots,X_9$ be the indicators of the events that the first, second, third, etc. cards are diamonds (in other words, for example, $X_3$ is $1$ if the third card is a diamond and $0$ otherwise). Then $X = \sum_{i=1}^9 X_i$, and by linearity of expectation the expectation $EX$ satisfies $EX = \sum_{i=1}^9EX_i$. Now, for every $i$ the probability that the $i$th card is a diamond is $1/4$. Therefore $EX = 9 \times 1/4 = 9/4$.
b) Assuming the multiples of three are $3,6,9$, each draw has probability $3/13$ of being a multiple of three. Using reasoning similar to the above the expected number of multiples of three is $4 \times 3/13 = 12/13$.
$\endgroup$ 2 $\begingroup$Let $X$ represent the number of diamonds drawn when randomly selected from a deck of cards.$$P(X=0)=(\frac{39}{52})(\frac{38}{51})...(\frac{32}{45})(\frac{31}{44})=\frac{21793}{378350}$$$$P(X=1)=\binom{9}{1}(\frac{39}{52})(\frac{38}{51})...(\frac{32}{45})(\frac{13}{44})=\frac{82251}{378350}$$$$P(X=2)=\binom{9}{2}(\frac{39}{52})(\frac{38}{51})...(\frac{33}{46})(\frac{13}{45})(\frac{12}{44})=\frac{246753}{756700}$$In general$$P(X=x)=\binom{9}{x}\frac{43!\cdot39!\cdot13!}{52!\cdot(30+x)!\cdot(13-x)!}$$So,$$E(X)=\sum_{x=0}^9 x\cdot P(X=x)=\sum_{x=0}^9 x\cdot\binom{9}{x}\frac{43!\cdot39!\cdot13!}{52!\cdot(30+x)!\cdot(13-x)!}=\frac{9}{4}$$After reading Micapps answer this seems stupid, but I'll post it as an answer anyway.
$\endgroup$
