prob of not happening

**sfspitfire23** · July 26th 2010, 10:59 AM

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?

**Plato** · July 26th 2010, 11:23 AM

Assuming independence (and we must), that is correct.

**Archie Meade** · July 26th 2010, 12:00 PM

Originally Posted by sfspitfire23

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?

"Neither" of them hitting the target...

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

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

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

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...

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

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