My daughter buys a \"magic” deck of cards, which consists of a standard deck of

Question

My daughter buys a "magic” deck of cards, which consists of a standard deck of cards, with x added AV (so there are (x+1) AV in total). For each part (a) to (d), make sure to name the distribution as part of your working. Each part is played with the same "magic" deck, so can use any information given in one part for the other parts. (a) I select a card, and replace it, ten times. The expected value of the number of A. (b) What is the probability I get exactly one Ay when I select a card and replace it (c) I play a different game with the same "magic” deck. (d) What is the probability I don't pull out an A my first three attempts? I will get is 3.2. What is the variance of the number of AI will get? ten times, given that I get less than two Av each time, until I get an AV. How many cards do I expect to pull out to get an A I pull out a card, replacing (e) What is the probability the second time I pull an AV is on my fifth attempt?

Explanation / Answer

Answer to part a)

Expected value = n*p

We got n = 10
EV = 3.2

Thus , 3.2 = 10*p

p = 0.32

.

Variance = n*p*(1-p)

Variance = 10 *0.32 *0.68 = 2.176

.

Answer to part b)

P(x < 2) = P(x=0) + P(x=1)

p = 0.32 , n = 10

P(x=0) = 10C0 *0.32^0 * 0.68^10 = 0.021

P(x=1) = 10C1 *0.32^1 * 0.68^9 = 0.099

P(x =1 | x < 2) = 0.099 / (0.021+0.099) = 0.825

.

Answer to part c)

Since the P value = 0.32 for A heart card

the number of cards need to pulled out is : 3.2, three cards minimum

.

Answer to part d)

P(no A heart in first three attempts) = (1-0.32)^3 = 0.314432

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