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Chapter 1: Problem 5

Let us select five cards at random and without replacement from an ordinary deck of playing cards. (a) Find the pmf of \(X\), the number of hearts in the five cards. (b) Determine \(P(X \leq 1)\).

Short Answer

Expert verified
The PMF is calculated using the formula \[p(x) = \frac{C(13, x) \cdot C(39, 5-x)}{C(52, 5)}\] where \(x\) ranges from 0 to 5. The probability of \(P(X \leq 1)\) is calculated by adding the individual probabilities when \(X = 0\) and \(X = 1\).

Step by step solution

01

Bucketing of the Cards

The ordinary deck has 52 cards which includes 4 suits (hearts, diamonds, clubs, spades) each with 13 cards. So, there are 13 hearts and 39 cards which are not hearts.
02

Calculate PMF

To find the PMF of \(X\), the number of hearts in the drawn cards, one needs to calculate the probability for all \(X = 0, 1, 2, 3, 4, 5\). In this case, one can use the concept of combination to find the number of ways to select \(X\) hearts from 13 hearts (\(C(13, X)\)) and (5 - \(X\)) non-hearts from 39 non-hearts (\(C(39, 5-X)\)). Then, one can find the PMF by dividing the product of the number of ways to select hearts and non-hearts by the total number of ways to select 5 cards from the deck (\(C(52, 5)\)).
03

Calculate \(P(X \leq 1)\)

This requires the sum of probabilities when \(X = 0\) and \(X = 1\). Applying the same combination method as previously, the probability when \(X = 0\) (no hearts) can be found through choosing 5 non-hearts from 39, divided by the total number of ways to select 5 cards. Meanwhile, when \(X = 1\), it implies 1 heart out of 13 and 4 non-hearts from 39. Their sum will give \(P(X \leq 1)\).

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Most popular questions from this chapter

A bowl contains 10 chips, of which 8 are marked \(\$ 2\) each and 2 are marked \(\$ 5\) each. Let a person choose, at random and without replacement, three chips from this bowl. If the person is to receive the sum of the resulting amounts, find his expectation.

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Let \(X\) have the pmf \(p(x)=1 / k, x=1,2,3, \ldots, k\), zero elsewhere. Show that the \(\mathrm{mgf}\) is $$ M(t)=\left\\{\begin{array}{ll} \frac{e^{t}\left(1-e^{k t}\right)}{k\left(1-e^{t}\right)} & t \neq 0 \\ 1 & t=0 \end{array}\right. $$

Let \(X\) be a random variable with a pdf \(f(x)\) and \(\operatorname{mgf} M(t)\). Suppose \(f\) is symmetric about \(0 ;\) i.e., \(f(-x)=f(x)\). Show that \(M(-t)=M(t)\).
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