Question

(1 point) Suppose that you randomly draw one card from a standard deck of 52 cards. After writing down which card was drawn, you replace the card, and draw another card. You repeat this process until you have drawn 15 cards in all. What is the probability of drawing at least 5 diamonds? For the experiment above, let X denote the number of diamonds that are drawn. For this random variable, find its expected value and standard deviation E(X)=( o=

Answer 1

Number of diamonds = 13

Total number of cards = 52

P[ a diamond is drawn ] = 12/52 = 1/4

Number of total cards drawn = 15

this is a case of binomial distribution with p = 0.25 and n = 15

P[ At least 5 diamond ] = 1 - P[ X <= 4 diamonds ]

Therefore, we get that

$\Pr(X \le 4) = \sum_{i=0}^{ 4} \Pr(X = i) = \sum_{i=0}^{ 4} \left( \begin{matrix} n \\ i \end{matrix}\right) p^i (1-p)^{n-i}$

This implies that

$\Pr(X \le 4) = \Pr(X = 0) + \Pr(X = 1) + \Pr(X = 2) + \Pr(X = 3) + \Pr(X = 4)$

$\Pr(X \le 4)= \left( \begin{matrix} 15 \\ 0 \end{matrix}\right) 0.25^{ 0} \times 0.75^{ 15-0} + \left( \begin{matrix} 15 \\ 1 \end{matrix}\right) 0.25^{ 1} \times 0.75^{ 15-1} + \left( \begin{matrix} 15 \\ 2 \end{matrix}\right) 0.25^{ 2} \times 0.75^{ 15-2} + \left( \begin{matrix} 15 \\ 3 \end{matrix}\right) 0.25^{ 3} \times 0.75^{ 15-3} + \left( \begin{matrix} 15 \\ 4 \end{matrix}\right) 0.25^{ 4} \times 0.75^{ 15-4}$

$\Pr(X \le 4)= 0.0134 + 0.0668 + 0.1559 + 0.2252 + 0.2252$

P[ X <= 4 diamonds ] = 0.6865

P[ At least 5 diamond ] = 1 - P[ X <= 4 diamonds ]

P[ At least 5 diamond ] = 1 - 0.6865

P[ At least 5 diamond ] = 0.3135

E(X) = np = 0.25*15 = 3.75

$\sigma = \sqrt{np(1-p)}$

$\sigma = \sqrt{15*0.25(1-0.25)}$

$\sigma = 1.6771$