I need help solving the problem.
1 answer:
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Answer:
a) 0.98% probability that 17 of the 29 people are men.
b) 3.86% probability that the first woman is reached on the 8th call
Step-by-step explanation:
For each person chosen by the agency, there are only two possible outcomes. Either it is a man, or it is a woman. The probability of selecting a man or a women in each trial is independent from other trials. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
(a) What is the probability that 17 of the 29 people are men?
This is P(X = 17) when . So
0.98% probability that 17 of the 29 people are men.
(b) What is the probability that the first woman is reached on the 8th call?
On the first 7 trials, all men, each with a 78% probability.
On the 8th trial, a women, with a 22% probability. So
3.86% probability that the first woman is reached on the 8th call
Using hit and trial, you van find it pretty easily-
- Putting x=0, we get y=-3
- Putting x=1, we get y=-1
- Putting x=2, we get y=5
You can try it for yourself as well, for me, I couldn't find any other option appropriate.