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

Given a non uniform sample space and the probabilities associated with the points, we defined the probability of an event A as the sum of the probabilities associated with the sample points favorable to A. [You used this definition in Problem 15with the sample space (2.5).] Show that this definition is consistent with the definition by equally likely cases if there is also a uniform sample space for the problem (as there was in Problem 15). Hint: Let the uniform sample space have n<Npoints each with the probability N-1. Let the nonuniform sample space have n points, the first point corresponding to N1 points of the uniform space, the second to N2 points, etc. What is N1 + N2 + .... Nn ?What are p1, p2, ...the probabilities associated with the first, second, etc., points of the nonuniform space? What is p1 + p2 +....+ pn? Now consider an event for which several points, say i, j, k, of the nonuniform sample space are favorable. Then using the nonuniform sample space, we have, by definition of the probability p of the event, p = pi + pj + pk . Write this in terms of the N’s and show that the result is the same as that obtained by equally likely cases using the uniform space. Refer to Problem 15as a specific example if you need to.

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Suppose a coin is tossed three times. Let x be a random variable whose value is 1 if the number of heads is divisible by 3, and 0 otherwise. Set up the sample space for x and the associated probabilities. Find x and σ.

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