17.0k views
0 votes
The probability of randomly selecting a heart or a queen is.

a) 1/13
b) 1/26
c) 1/52
d) 1/104

User Vimalnath
by
8.4k points

1 Answer

4 votes

Final answer:

The probability of drawing a heart or a queen from a standard deck of cards is 15 out of 52, as we need to count all the hearts and the remaining queens except the queen of hearts.

Step-by-step explanation:

The question asks for the probability of randomly selecting a heart or a queen from a standard deck of playing cards. In a standard deck, there are 13 hearts and 4 queens. However, since one of the queens is also a heart (the queen of hearts), there are 12 other hearts and 3 other queens (making a total of 16 possible cards) that meet the criteria without counting any card twice. Therefore, to find the probability, we add the probability of selecting a heart (13 out of 52) to the probability of selecting a queen from the other three suits (3 out of 52) and subtract the overlap which is the queen of hearts (1 out of 52)Probability = (Number of hearts + Number of queens from other suits - Queen of hearts) / Total number of cardsProbability = (13 + 3 - 1) / 5Probability = 15 / The probability of randomly selecting a heart or a queen can be calculated by adding the probabilities of selecting a heart and selecting a queen, then subtracting the

probability of selecting both a heart and a queen since they are not mutually exclusive events.There are 13 hearts in a deck of 52 cards, so the probability of selecting a heart is 13/52.There are 4 queens in a deck of 52 cards, so the probability of selecting a queen is 4/52Since there is 1 queen of hearts in a deck of 52 cards, the probability of selecting both a heart and a queen is 1/52To find the probability of selecting a heart or a queen, we can use the formulaP(heart or queen) = P(heart) + P(queen) - P(heart and queen)P(heart or queen) = (13/52) + (4/52) - (1/52Simplifying the expression gives:P(heart or queen) = 16/52The probability simplifies to 15 divided by 52, which can be reduced to approximately 0.2885 or 1 in 3.47, which is not represented by any of the provided options (a) 1/13, (b) 1/26, (c) 1/52, (d) 1/104. Hence, it seems there may have been a mistake in the question or the options provided.

User Rick Regan
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.