SOLUTION: 1) There are 10 people in the room. 6 are male and 4 are female. 5 of them have been arrested at some point in their lives. 4 of these five people are male. If you select 5 people
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-> SOLUTION: 1) There are 10 people in the room. 6 are male and 4 are female. 5 of them have been arrested at some point in their lives. 4 of these five people are male. If you select 5 people
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Question 31444: 1) There are 10 people in the room. 6 are male and 4 are female. 5 of them have been arrested at some point in their lives. 4 of these five people are male. If you select 5 people at random what is the probability that at least 3 of them are males who have been arrested and at least one of them is a female who has not been arrested? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 1) There are 10 people in the room. 6 are male and 4 are female. 5 of them have been arrested at some point in their lives. 4 of these five people are male. If you select 5 people at random what is the probability that at least 3 of them are males who have been arrested and at least one of them is a female who has not been arrested?
Draw the Venn diagram of your problem.
Your selection requirements can be met in two
disjoint ways:
Case 1: Select 3 arrested males from the four available and
select 2 females from the 3 that have not been arrested.
Number of ways to do this is 4C3*3C2 = 4(3)= 12
Case 2: Select 4 arrested males from the four available and
select 1 female from the 3 that have not been arrested.
Number of ways to do this is 4C4*3C1 = 1(3)=3
Total number of ways to succeed is 12+3=15
Total number of groups of 5 that could be selected is 10C5
Probability of meeting your selection criteria = 16/10C5 = 0.0595...
Cheers,
Stan H.