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jerryragland
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This is from the resource section. If p is the probability of the event happening then the probability of the event not happening is 1-P. According to this, I was deriving. P(Missing every time) = 1 - P(hitting every time). Is P(Missing every time) and P(Missing at least once) both are one and same. May be an English question...
Example 6
A bowman hits his target in 1/2 of his shots. What is the probability of him missing the target at least once in three shots?
Solution
An optimal way to solve this is to think that (missing the target at least once) = 1 - (hitting it every time). So, p(hitting it every time) = p(shot1_hit and shot2_hit and shot3_hit) = p(shot1_hit) * p(shot2_hit) * p(shot3_hit) = 1/2 * 1/2 * 1/2 = 1/8; p(missing at least once) = 1 - p(hitting it every time) = 1 - 1/8 = 7/8.
Alternatively, use the F/T rule. The T are HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM. T = 8. The F are HHM, HMH, HMM, MHH, MHM, MMH, MMM. F = 7.
In cases like this it is evident that F/T rule soon becomes too hard to apply.
Example 6
A bowman hits his target in 1/2 of his shots. What is the probability of him missing the target at least once in three shots?
Solution
An optimal way to solve this is to think that (missing the target at least once) = 1 - (hitting it every time). So, p(hitting it every time) = p(shot1_hit and shot2_hit and shot3_hit) = p(shot1_hit) * p(shot2_hit) * p(shot3_hit) = 1/2 * 1/2 * 1/2 = 1/8; p(missing at least once) = 1 - p(hitting it every time) = 1 - 1/8 = 7/8.
Alternatively, use the F/T rule. The T are HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM. T = 8. The F are HHM, HMH, HMM, MHH, MHM, MMH, MMM. F = 7.
In cases like this it is evident that F/T rule soon becomes too hard to apply.
