Question

1. Two cards are selected at random from a standard deck of 52 cards without replacement. What is the probability that both cards are diamonds? Submit your answer as a decimal rounded to the nearest thousandth.

1. A jar contains 14 blue candies, 10 green candies, and 5 yellow candies. Three candies are selected at random without replacement. What is the probability that the first is yellow, the second is blue, and the third is yellow? Submit your answer as a decimal rounded to the nearest thousandth.

1. In a group of 400 employees (142 men and 258 women), 29 of the men and 39 of the women work in accounting.  If an accountant is selected at random, what is the probability the accountant is a woman? Round your answer to the nearest thousandth.

1. In a group of 400 employees (142 men and 258 women), 29 of the men and 39 of the women work in accounting. If a male employee is selected at random, what is the probability that he is an accountant? Round your answer to the nearest thousandth.

Answer 1

Solved 2 questions (Note: allowed to solve 1 question per post)

Two cards are selected at random from a standard deck of 52 cards without replacement. What is the probability that both cards are diamonds? Submit your answer as a decimal rounded to the nearest thousandth.

In each deck, there are 52 cards and 13 cards of each type.(diamonds, hearts etc)

No. ways in which we can pull the 1st card = 13/52

Since the second card is drawn without replacement

No. ways in which we can pull the 2nd card = 12/51

P(both cards are diamond= (13/52)*(12/51)= 0.0588

A jar contains 14 blue candies, 10 green candies, and 5 yellow candies. Three candies are selected at random without replacement. What is the probability that the first is yellow, the second is blue, and the third is yellow? Submit your answer as a decimal rounded to the nearest thousandth.

Total number of candies = 14+10 + 5 = 29

Blue = 14
Green = 10
yellow = 5

Probability of selecting the first as a yellow candy = 5/29

Since we do it without replacement we have 29 candies in the jar.

Probability of selecting the second as a blue candy = 14/28

Probability of selecting the third as a blue candy = 4/27

Note : We have taken one yellow in the first draw, hence we have 4 yellow remaining

Need probability = ( 5/29)*(14/28)*(4/27)= 0.012771392