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Question:
A man opens a general store and a book shop. In any one year the probability of a robbery in the general store is 0.10 and the probability of a robbery in the book shop is 0.02. For any one year. What is the probability that:1)Neither business will be robbed.
2)Either one or the other (but not both) will be robbed
Answer
a. (0.90)(0.98) b. 0.10 + 0.02 - (0.10)(0.02)
Or
Solution:-
Probability equals the number of success divided by the total.
The total probability is always equal to 1.
Probability of success is equal to 0.10
This means the probability of failure equals 0.90
Subtract 0.10 from 1 and you get 0.90.
Odds for equal the number of successes divided by the number of failures.
Odds of success are therefore equal to 0.10 /0.90
Odds against equal the number of failure divided by the number of successes.
Odds of failure are therefore equal to 0.90 / 0.10
If you let p = the probability of success and you let q = the probabilities of failure, then you get:
Probability of success = p / (p + q). {0.10/ (0.10+0.90)} =0.90
Probability of failure = q / (p + q). {0.90/ (0.10+0.90)} =0.10
Similarly
The probability of a robbery in the book shop is 0.02.
Probability of success is equal to 0.02
This means the probability of failure equals 0.98
Probability of success = p / (p + q). {0.02/ (0.02+0.98)} =0.98
Probability of failure = q / (p + q). {0.98/ (0.02+0.98)} =0.02
Neither business will be robbed
(0.90)(0.98) probabilities of failure
Either one or the other (but not both) will be robbed.
0.10 + 0.02 - (0.10)(0.02)
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