Suppose you are repeatedly picking one random card from a deck of cards with replacement (that means, you pick a card randomly, see it, put the card back to the deck again and shuffle. You keep doing it repeatedly )

i. What is the probability that you get your first Diamonds card on the 3rd pick ?

ii. How many times are you expected to pick a card to get a KING ?

iii. What is the probability that you'll get exactly 3 spades after picking 5 times?

iv. What is the probability that you'll get AT MOST 2 clubs after picking 5 times ?

 

i. 1/4 or 0.25

ii. 13 times

iii. 0.0879

iv. 0.8965

Step-by-step explanation

i. There are 13 diamonds in a 52 deck of cards. Picking a card randomly, see it, put the card back to the deck again and shuffle, the probability doesn't change whether on 1st pick or 3rd pick.

Probability = 13/52 = 1/4 or 0.25

ii. There are 4 kings in a 52 deck of cards.

Probability = 4/52 = 1/13

To get a king, it's probability must be equal to 1, so let x as times you expected to pick a card to get a KING.

1 = 1/13 * x

x = 13 times

iii. There are 13 spades in a 52 deck of cards.

Probability, p = 13/52 = 0.25

Using the binomial probability P = Cⁿ_xp^xq^n-x = ( n!/(n-x)!x! ) * p^xq^n-x ,

n = 5 ; x = 3 ; q = 1 - p = 1 - 1/4 = 0.75

P(x=3) = ( 5! / ( 5 - 3 )!3! ) * 0.25³ * 0.75² = 0.0879

iv. There are 13 clubs in a 52 deck of cards.

robability, p = 13/52 = 0.25

Using the binomial probability P = Cⁿ_xp^xq^n-x = ( n!/(n-x)!x! ) * p^xq^n-x ,

n = 5 ; x ≤ 2 ; q = 1 - p = 1 - 1/4 = 0.75

P(x≤ 2 ) = P(x=2) + P(x=1) + P(x=0)

P(x≤ 2 ) = [( 5! / ( 5 - 2 )!2! ) * 0.25² * 0.75³ ] + [( 5! / ( 5 - 2 )!2! ) * 0.25² * 0.75³ ] + [( 5! / ( 5 - 2 )!2! ) * 0.25² * 0.75³ ]

P(x≤ 2 ) = 0.2637 + 0.3955 + 0.2373 = 0.8965