Question

A standard deck of cards contains 52 cards. One card is selected from the deck. ​(a) Compute the probability of randomly selecting a jack or king. ​(b) Compute the probability of randomly selecting a jack or king or nine. ​(c) Compute the probability of randomly selecting a two or spade.

Answer 1

Solution:

Given: A standard deck of cards contains 52 cards. One card is selected from the deck.

n = total number of ways in which we can select any one card out of 52 cards = ₅₂C₁ = 52 ways

Where

Part a) Compute the probability of randomly selecting a jack or king.

That is: P( A Jack or a King card) = ..........?

Since Jack and King card are mutually exclusive , we can write:

P( A Jack or a King card) = P(a Jack) + P(a King)

there are 4 jack cards and 4 king cards,

thus one Jack card from 4 Jack can be selected in 4C1 = 4 ways

similarly one King card from 4 King can be selected in 4C1 = 4 ways

where

Thus

P( A Jack or a King card) = P(a Jack) + P(a King)

P( A Jack or a King card) = 4/52 + 4/52

P( A Jack or a King card) = (4+4) / 52

P( A Jack or a King card) = 8/52

P( A Jack or a King card) = 2/13

P( A Jack or a King card) = 0.153846

Part b) Compute the probability of randomly selecting a jack or king or nine.

P( A king or a Nine) = .......?

P( A king or a Nine) = P(King) + P(Nine)

one King card from 4 King can be selected in 4C1 = 4 ways

similarly one "Nine" card from 4 "Nine" can be selected in 4C1 = 4 ways

Thus

P( A king or a Nine) = P(King) + P(Nine)

P( A king or a Nine) = 4/52 + 4/52

P( A king or a Nine) = (4+4) / 52

P( A king or a Nine) = 8 / 52

P( A king or a Nine) = 2 / 13

P( A king or a Nine) = 0.153846

Part c) Compute the probability of randomly selecting a two or spade.

P( A "Two" or a Spade) =.........?

Since one of Two's is a spade , Two and Spade are not mutually exclusive

thus we use addition rule of probability:

P(A or B) = P(A) + P(B) - P( A and B)

Thus

P( A "Two" or a Spade) = P( A "Two") + P( a Spade) - P( A "Two" and a Spade)

one "Two" card from 4 "Two" can be selected in ₄C₁ = 4 ways

We have 13 spade cards

Thus a Spade card can be selected in ₁₃C₁ = 13 ways

where

and

there is a one Spade card which is number "Two"

Thus there is only way of selecting a Two and a Spade

Thus

P( A "Two" or a Spade) = P( A "Two") + P( a Spade) - P( A "Two" and a Spade)

P( A "Two" or a Spade) = 4/52 + 13/52 - 1/52

P( A "Two" or a Spade) = (4+13-1) / 52

P( A "Two" or a Spade) = 16 / 52

P( A "Two" or a Spade) = 4 / 13

P( A "Two" or a Spade) = 0.307692