Given ‘A‘s’ probability of hitting a target is 2/3, ‘B’s’ probability of hitting the same target is 4/7. What is the probability of not hitting the target by either of them?

I get 1/7. Is this correct?

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- Jul 26th 2010, 10:59 AMsfspitfire23prob of not happening
**Given ‘A‘s’ probability of hitting a target is 2/3, ‘B’s’ probability of hitting the same target is 4/7. What is the probability of not hitting the target by either of them?**

I get 1/7. Is this correct?

- Jul 26th 2010, 11:23 AMPlato
Assuming independence (and we must), that is correct.

- Jul 26th 2010, 12:00 PMArchie Meade
"Neither" of them hitting the target...

One way is..... $\displaystyle P(both\ miss)=P(A\ misses)P(B\ misses)=\frac{1}{3}\frac{3}{7}$

Another way is....$\displaystyle 1-P(both\ hit)-P(A\ hits\ and\ B\ misses)-P(B\ hits\ and\ A\ misses)$

$\displaystyle =1-\frac{2}{3}\frac{4}{7}-\frac{2}{3}\frac{3}{7}-\frac{4}{7}\frac{1}{7}=1-\frac{18}{21}$

A possible interpretation of the target not being hit by "either" of them is

"the probability of the target not being hit by either one of them" which is

"the probability of A or B missing"

which would be...

$\displaystyle P(A\ hits\ and\ B\ misses)+P(A\ misses\ and\ B\ hits)+P(both\ miss)$