SOLUTION: Table 3: Survey Results Male Female Total Less than one month’s income 66 83 149 One month’s income or more 76 62 138 Total 14

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Question 1142588: Table 3: Survey Results
Male Female Total
Less than one month’s income 66 83 149
One month’s income or more 76 62 138
Total 142 145 287
Three separate workers are selected at random. Find the probability that at least one of the three workers has one month’s income or more set aside for emergencies:
(this is what I have so far)
P(1st worker has one month's income) = 138/287= 0.481
P(1st worker has one month's income | 2nd worker has one month's income) = 137/286= 0.479
P(1st worker has one month's income | 3rd worker has one month's income )= 136/285=0.4777

Answer by ikleyn(52903) (Show Source):
You can put this solution on YOUR website!
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You are given the Table Table 3: Survey Results Male Female Total Less than one month’s income 66 83 149 One month’s income or more 76 62 138 Total 142 145 287 Three separate workers are selected at random. Find the probability that at least one of the three workers has one month’s income or more set aside for emergencies:

I will solve the problem by MY WAY and then you will have an opportunity to compare my result with yours.

The question is Find the probability that at least one of the three workers has one month’s income or more. I will consider the COMPLEMENT probability that no one of the three randomly selected workers has one month’s income or more. This complement probability is . Hence, the answer to the problem's question is = 0.8614 (approximately) = 86.14%.