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Chapter 2: Q.2.18 (page 49)

Two cards are randomly selected from an ordinary playing deck. What is the probability that they form a blackjack? That is, what is the probability that one of the cards is an ace and the other one is either a ten, a jack, a

queen, or a king?

Short Answer

Expert verified

We have0.048///0.0965.

Step by step solution

01

Given Information.

Two cards are randomly selected from an ordinary playing deck.

02

Explanation

P(blackjack)=(4Cl)*(16Cl)/(52C2)=(4*16)/(1326)=0.048

03

Explanation.The total number of ten, jack, queen, and king in a deck of cards is equal to 4+4+4+4=16. Choosing 1 card apart from 1 ace out of 16 possible cards is equal to 16C1. Further,  there are  4 aces in a deck of cards. Selecting 1 ace from 4 possible aces is equal to 4C1. Also, there are two positions of a selection of cards which means an ace can be either at the first position or last. That is, the ordering also important.

P(1aceand1oneiseitheraten,ajack,aqueen,oraqueen)=(4Cl)*(16Cl)*2!/(52C2)=(4162)/(1326)=0.0965

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Most popular questions from this chapter

A woman has n keys, of which one will open her door.

(a) If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her kth try?

(b) What if she does not discard previously tried keys?

Two symmetric dice have had two of their sides painted red, two sides painted black, one painted yellow and the other painted white. When this pair of dice rolled, what is the probability that both dice land with same color face up?

A pair of dice is rolled until a sum of either 5or 7appears. Find the probability that a 5occurs first.

Hint: Let Endenote the event that a 5occurs on the nth roll and no 5or7occurs on the first n-1rolls. Compute P(En) and argue that ∑n=1∞P(En)is the desired probability

Use Venn diagrams

(a)to simplify the expressions (E∪F)(E∪Fc);

(b)to prove DeMorgan’s laws for eventsEandF. [That is, prove(E∪F)c=EcFc, and (EF)c=Ec∪Fc]

Consider Example5o, which is concerned with the number of runs of wins obtained whennwins and mlosses are randomly permuted. Now consider the total number of runs—that is, win runs plus loss runs—and show that

P(2kruns)=n-1k-1m-1k-1n+mnP(2k+1runs)=n-1km-1k-1+n-1k-1m-1kn+mn
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