A card is selected from a shuffled deck. What is the probability that it is either a king or a club? That it is both a king and a club?

Short Answer

Expert verified

The probability that the card is either a king and a club is 1/52 and the probability that the number is either a king and a club or both is 4/13.

Step by step solution

01

Expression for the probability that either one of the events will occur

The expression for the probability that either one of the events will occur is as follows,

P(A + B) = P(A) + P(B) -P(AB)

Here, P(A) is one event, and P(B) is another event.

02

Determination of the probability that the card is either a king and a club

In a deck of card, there are 52 cards of 4 suits namely spade, club, diamond and heart, out of which spade and club are black and diamond and heart are red.

There are 13 cards of each suit and 26 cards of each color.

Let A represents the event that card is king and B represents the event that card is club.

In a deck, there are 13 club cards, out of which only one is king, So, the probability is as follows,

P(AB) = 1/52

Thus, the probability that the number is either a king and a club or both is 1/52.

03

Determination of the probability that the card is either a king and a club or both

Write the probability that there are 4 king cards.

P(A) = 4/52

Write the probability that there are 13 club cards.

P(B) = 13/52

Write the expression for the probability that the card is either a king and a club or both.

P(A + B) = P(A) + P(B) -P(AB)

Substitute all the values in the above expression.

P (A + B) = (4/52) + (13/52) - (1/52)

= 16/52

= 4/13

Thus, the probability that the card is either a king and a club or both is 4/13.

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Most popular questions from this chapter

(a) A weighted coin has probability of 23¯of showing heads and 13of showing tails. Find the probabilities of in two tosses of the coin. Set up the sample space and the associated probabilities. Do the probabilities add to 1 as they should? What is the probability of at least one head? What is the probability of two heads if you know there was at least one head?

(b) For the coin in (a), set up the sample space for three tosses, find the associated probabilities, and use it to answer the questions in Problem 2.12.

Define s by the equation.s2=(1/n)i=1n(xix¯)2Show that the expected valueof.s2is[(n1)/n]σ2Hints: Write

(xix¯)2=[(xiμ)(x¯μ)]2=(xiμ)22(xiμ)(x¯μ)+(x¯μ)2

Find the average value of the first term from the definition ofσ2and the average value of the third term from Problem 2. To find the average value of the middle term write

(x¯μ)=(x1+x2++xnnμ)=1n[(x1μ)+(x2μ)++(xnμ)]

Show by Problemthat

E[(xiμ)(xjμ)]=E(xiμ)E(xjμ)=0   forij

andevaluate6.14 (same as the first term). Collect terms to find

E(s2)=n1nσ2

Find the number of ways of putting 2 particles in 5 boxes according to the different kinds of statistics.

Question: Use both the sample space (2.4) and the sample space (2.5) to answer the following questions about a toss of two dice.

(a) What is the probability that the sum is ≥ 4?

(b) What is the probability that the sum is even?

(c) What is the probability that the sum is divisible by 3?

(d) If the sum is odd, what is the probability that it is equal to 7?

(e) What is the probability that the product of the numbers on the two dice is 12?

(a) Repeat Problem 6 where the “circular” area is now on the curved surface of the earth, say all points at distance s from Chicago (measured along a great circle on the earth’s surface) with sπR3where R = radius of the earth. The seeds could be replaced by, say, radioactive fallout particles (assuming these to be uniformly distributed over the surface of the earth). Find F(s)andf(s) .

(b) Also find F(s)andf(s) ifs1<<R (say s1mile where R=4000miles). Do your answers then reduce to those in Problem 6?

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