A trick deck of cards is printed with the hearts and diamonds black, and the spades and clubs red. A card is chosen at random from this deck (after it is shuffled). Find the probability that it is either a red card or the queen of hearts. That it is either a red face card or a club. That it is either a red ace or a diamond.

Short Answer

Expert verified

The probability that the drawn card is either a red card or the queen of hearts is 27/52.

The probability chosen card either a red face card or a club is 4/13.

The probability the chosen card is either a red ace or a diamond is 15/52.

Step by step solution

01

Definition of Probability

The probability of any event is determined by dividing the number of favorable outcomes of a particular even to the total number of outcomes. Using this techniques probability of any event can be determined.

02

Determination of the probability that the chosen card is either a red card or the queen of hearts

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.

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

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

In this deck of card, there is 1 queen of hearts (which is no red) and 26 red cards, this implies that total number of possibilities for a card to be a queen of heart or red is as follows,

1+26=27

This implies that the number of outcomes favorable are 27 and total number of outcomes are 52.

Write the expression for the probability.

P = number of outcomes favorable to E/total number of outcomes ...(i) Substitute the values in the above expression.

P = 27/52

03

Determination of the probability that the chosen card either a red face card or a club 

In a deck of card, there are 13 club cards (which are red) and 6 red face cards out of which 3 are of clubs, this implies that total number of possibilities for a card to be a red face card or a club is 13 + 6 - 3 = 16.

This implies that the number of outcomes favorable are 16 and total number of outcomes are 52.

Substitute the values in the equation (i) to find the probability chosen card either a red face card or a club.

P = 16/52

= 4/13

04

Determination of the probability that the chosen card is either a red ace or a diamond

In this deck of card, there are 2 red aces and 13 diamond cards, this implies that total number of possibilities for a card to be either a red ace or a diamond is 2 + 13 = 15

This implies that the number of outcomes favorable are 15 and total number of outcomes are 52.

Substitute the values in the equation (i) to find the probability the chosen card is either a red ace or a diamond.

P = 15/52

Thus, the probability that the drawn card is either a red card or the queen of hearts is 27/52, the probability chosen card either a red face card or a club is 4/13, and the probability the chosen card is either a red ace or a diamond is 15/52.

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Most popular questions from this chapter

(a) In Example, 5a mathematical model is discussed which claims to give a distribution of identical balls into boxes in such a way that all distinguishable arrangements are equally probable (Bose-Einstein statistics). Prove this by showing that the probability of a distribution of N balls into n boxes (according to this model) with N1 balls in the first box, N2in the second, ··· , Nn in thenth , is1C(n1+N,N) for any set of numbers Ni such thatNii=1nNi=N.

b) Show that the model in (a) leads to Maxwell-Boltzmann statistics if the drawn card is replaced (but no extra card added) and to Fermi-Dirac statistics if the drawn card is not replaced. Hint: Calculate in each case the number of possible arrangements of the balls in the boxes. First do the problem of 4particles in 6boxes as in the example, and then do N particles in n boxes (n>N ) to get the results in Problem19 .

Two dice are thrown. Given the information that the number on the first die iseven, and the number on the second is <4, set up an appropriate sample space andanswer the following questions.

(a) What are the possible sums and their probabilities?

(b) What is the most probable sum?

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

Use the sample space of Example 1 above, or one or more of your sample spaces in Problem 11, to answer the following questions.

(a) If there were more heads than tails, what is the probability of one tail?

(b) If two heads did not appear in succession, what is the probability of all tails?

(c) If the coins did not all fall alike, what is the probability that two in succession

were alike?

(d) If Nt=numberoftailsand Nh=numberofheads, what is the probability

That |Nh-Nt|=1?

(e) If there was at least one head, what is the probability of exactly two heads?

Computer plot on the same axes the normal density functions with μ=0σ=1and, 2, and 5. Label each curve with itsσ.

Are the following correct non-uniform sample spaces for a throw of two dice? If

so, find the probabilities of the given sample points. If not show what is wrong.

Suggestion: Copy sample space (2.4) and circle on it the regions corresponding to the points of the proposed non-uniform spaces.

(a) First die shows an even number.

First die shows an odd number.

(b) Sum of two numbers on dice is even.

First die is even and second odd.

First die is odd and second even.

(c) First die shows a number≤3.

At least one die shows a number > 3.

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