Question

3. A hand of five cards is drawn simultaneously (without order or replacement) from a standard 52-card deck. Let A be the event that the hand includes four cards of the same kind, and let B be the event that at least two of the cards in the hand are of the same kind. a. Compute P(A) (5) b. Compute P(B). (5) c. Compute P(AB) (5) 4. Suppose Z and X are continuous random variables such that Z has a standard normal distribution and X = 52 + 10. a. Compute P(7 < X < 17). [6] b. What are the expected value E(X) and variance V(X) of X? [6] c. What kind of distribution does X have? (3)

Answer 1

3 )    Let A be the event that the hand includes four cards of the same kind

P(drawing a card of one kind) = 13/52                                 , since there 13 cards of kind in a total of 52 cards.

P(drawing second card of same kind) = 12/51                 ,since it is without replacement, now there is total 51 cards                                                                                          and 12 of the same kind left. one is already taken)

simillarly, P(Drawing 3rd card of same kind)=11/50

P(Drawing 4th card of same kind)=10/49

therefore P(drawing 4 cards of same kind) = $\tiny \frac{13\times 12\times 11\times 10}{52\times51\times 50\times 49 }$ = 17160/6497400 =0.00264

In a more simple way, we can say

P(A) = P(Drawing 4 cards of same kind)=$\tiny \frac{combination\, of\, 4\, cards\, from \, 13\, cards\, of\, same \, kind }{total \, \, combination \, of \, 4 \, cards \, from \, 52 \, cards}\,$   = $\tiny \frac{\binom{13}{4}}{\binom{52}{4}}$ =0.00264

P(B) =P(atleast two of the cards same) =$\tiny \frac{13}{52}\times \frac{12}{51}\times \frac{50}{50}\times \frac{49}{49}$      = $\tiny \frac{\binom{13}{2}\times \binom{50}{2}}{\binom{52}{4}}$ = $\tiny \frac{13}{52}\times \frac{12}{51}$ = 0.05882

P(A/B) =$\tiny \frac{P(A\cap B)}{P(B)}$ =    $\tiny \frac{P(A)}{P(B)}$ = 0.044882

Since$\tiny A\cap B$ is the event hands include 4 cards of same kind and atleast 2 of same kind.

which implies the event hand includes 4 cards of same kind

i.e, $\tiny {P(A\cap B)$ =P(A)

4)   X=5Z+10

     X-10 =5Z

   Z=$\tiny \frac{X-10}{5}$

a) We have to find P(7$\tiny \leq$X$\tiny \leq$17). We have the table of standard normal distribution. therfore we convert this into Z.

P(7$\tiny \leq$X$\tiny \leq$17) = P(7-10$\tiny \leq$ X-10 $\tiny \leq$ 17-10) = $\tiny P(\frac{7-10}{5}\leq \frac{X-10}{5}\leq \frac{17-10}{5})$ = $\tiny P(-0.6 \leq Z \leq 1.4)$

                      =P(Z$\tiny \leq$1.4) - P(Z$\tiny \leq$ -0.6)    [ The figure below(3rd figure) shows the given area]

                     = 0.9192 -0.2743      from z table

                     =0.6449

23 1.4 area below 1.4 area below -0.6 -0.6 0 1.4

b) E(aX+b) = aE(X)+b

   V(aX+b) = a²V(X)           since variance of constant is zero.

E(X) = E(5Z+10) = 5E(Z) +10                 [    Z $\small \sim N(0,1)$ Therefore, E(Z) =0 and V(Z)=1 ]                                         =5*0+10 = 10

V(X)= V(5Z+10) = 25V(X) =25*1 =25

c)    If Y follows normal distribution, c be a constant , then cY and Y+c also follows normal.

Therefore, X=5Z+10 has normal distribution with mean 10 and variance 25.