There are 34 men and 32 women at a party. Of these, 13 men and 19 woman are married. If 2 people are chosen at random, find the probability that both will be men.
Answer: 17/65
Thanks for any help
Number of possible pairs chosen is 2 from 66
= C(66,2) = 2145
number of pairs consisting of men
= C(34,2) = 561
prob(that your pair is made up of 2 men)
= 561/2145 = 17/65
The information that 13 men and 19 women are married is irrelevant.
That makes so much sense. Thank you
Or, consider this as a problem of drawing without replacement:
P(man,man) = 34/66 * 33/65 = 1122/4290 = 17/65
To find the probability that both people chosen will be men, we need to calculate the ratio of the number of favorable outcomes (choosing 2 men) to the number of possible outcomes (choosing any 2 people).
Let's break down the problem step by step:
Step 1: Calculate the total number of people at the party.
Given that there are 34 men and 32 women, the total number of people is 34 + 32 = 66.
Step 2: Determine the number of favorable outcomes (choosing 2 men).
Since we want both people to be men, we need to choose 2 men out of the 34 available men. This can be calculated using the combination formula, often denoted as "nCr" or "C":
nCr = n! / [(n-r)! * r!]
In this case, we want to find 34C2:
34C2 = 34! / [(34-2)! * 2!]
= 34! / (32! * 2!)
Step 3: Calculate the number of possible outcomes (choosing any 2 people).
Since we want to choose any 2 people, regardless of their gender, we need to calculate 66C2:
66C2 = 66! / [(66-2)! * 2!]
= 66! / (64! * 2!)
Step 4: Calculate the probability.
The probability is the ratio of the number of favorable outcomes to the number of possible outcomes. Therefore:
Probability = (34C2) / (66C2)
Now let's substitute the values and calculate the probability:
Probability = (34! / (32! * 2!)) / (66! / (64! * 2!))
= [34(33)] / [66(65)]
= 1122 / 4290
= 17/65
Therefore, the probability that both people chosen will be men is 17/65.
I hope this explanation helps! Let me know if you have any further questions.