Binominal question....quick!!!

[匿名]

2013-02-08 5:11 am

I fires 9 shots at a target and each shot I hit the target with probability 0.7. Assume
that the shots are independent find the following:
a) Given that I hit the target at least twice, what is the probability that I hit the target
exactly 4 times?
b) Given that the first 2 shots I hit the target, what is the probability that I hit the
target exactly 4 times in the 9 shots?

更新1:

For a) P(hit the target exactly 4 times and hit the target at least twice)/ P(hit the target at least twice) = P(hit the target exactly 4 times | hit the target at least twice).

更新2:

Why we don't need to consider "and hit the target at least twice" and use the multiplication rule in the numerator?

更新3:

Same in b) P(hit the target exactly 4 times in the 9 shots and the first 2 shots I hit the target) / P(first 2 shots hit the target) = P(hit the target at the first 2 shots and exactly 2 times at other shots) / P(hit the target at the first 2 shots). Why?

更新4:

Thx!!!

回答 (1)

用戶2053

2013-02-08 7:38 am

✔ 最佳答案

a)P(hit the target exactly 4 times | hit the target at least twice)
= P(hit the target exactly 4 times | 1 - hit the target 0 or once)
= 9C4 * 0.7⁴* 0.3⁵ / (1 - 0.3⁹- 9C1 * 0.7 * 0.3⁸)
= 0.073545665
b)P(hit the target exactly 4 times | hit the target at the first 2 shots)= P(hit the target at the first 2 shots and exactly 2 times at other shots)
─────────────────────────────────────────
P(hit the target at the first 2 shots)
= 0.7² * 7C2 * 0.7² * 0.3⁵ / 0.7²= 0.0250047

2013-02-08 13:12:03 補充：
For a) ,
If hit the target exactly 4 times , then it must be hit the target at least twice.
(4 times already means at least twice)

But for b) ,
If hit the target exactly 4 times ,
then may not be hit the target at the first 2 shots.
So we need to consider more.

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