[MOVED] Probability: Five men and 4 women play in a baseball team.

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jramirez25

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How do you answer this question. Five men and 4 women play in a baseball team. Their names are put into a box and two persons are randomly selected to co-captain the team.
a. Calculate the probability that there will be at least 1 female captain.
b. Jay is one of the men on the team. He believes that he has a good chance of being one of the co-
captains. What is Jay probability to be one of the co-captains?
 
jramirez25 said:
How do you answer this question. Five men and 4 women play in a baseball team. Their names are put into a box and two persons are randomly selected to co-captain the team.
a. Calculate the probability that there will be at least 1 female captain.
b. Jay is one of the men on the team. He believes that he has a good chance of being one of the co-
captains. What is Jay probability to be one of the co-captains?
Click to expand...
That's pretty straightforward.

There are 9 people on the team, 5 men, 4 women. What is the probability that the first person drawn is a woman? Since we are asked for the probabliity of "at least one woman", what happens with the second draw is irrelevant.

What is the probablity the first person drawn is a man? If that happens there are 8 people left, 4 men and 4 women. What is the probability the second person drawn is a woman?

(Why was this question posted under "news"?)
 
Last edited:
jramirez25 said:
How do you answer this question. Five men and 4 women play in a baseball team. Their names are put into a box and two persons are randomly selected to co-captain the team.
a. Calculate the probability that there will be at least 1 female captain.
b. Jay is one of the men on the team. He believes that he has a good chance of being one of the co-
captains. What is Jay probability to be one of the co-captains?
Click to expand...
Quite often it is easier to answer a question that asks "at least one" by calculating probability of "none" and subtracting thet from unity:

\(\displaystyle P(\text{at least 1}) = 1 - P(\text{none})\)

After you have worked the problem following HallsofIvy,s approach, go back and do it this way:

\(\displaystyle P(\text{at least 1 woman}) = 1 - P(\text{1st not a woman}) \times P(\text{2nd not a woman})\)


I would recommend that same procedure for 1st is not Jay and 2nd is not Jay.
 
Probability

DrPhil said:
Quite often it is easier to answer a question that asks "at least one" by calculating probability of "none" and subtracting thet from unity:

\(\displaystyle P(\text{at least 1}) = 1 - P(\text{none})\)

After you have worked the problem following HallsofIvy,s approach, go back and do it this way:

\(\displaystyle P(\text{at least 1 woman}) = 1 - P(\text{1st not a woman}) \times P(\text{2nd not a woman})\)


I would recommend that same procedure for 1st is not Jay and 2nd is not Jay.
Click to expand...

Does it mean that 1-(8/9x7/8)=2/9
 
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