Shooting Together

Probability Level 2

5 7 \frac{5}{7} 6 7 \frac{6}{7} 16 21 \frac{16}{21} 17 21 \frac{17}{21}

Kumar Gupta by Kumar Gupta , 23, India

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3 solutions

Daniel Liu
Mar 3, 2014

Note that the probability that none of them hit the target is ( 1 − 1 3 ) ( 1 − 5 7 ) = 4 21 \left(1-\dfrac{1}{3}\right)\left(1-\dfrac{5}{7}\right)=\dfrac{4}{21} . Therefore the probability of at least one of them hitting the target is 1 − 4 21 = 17 21 1-\dfrac{4}{21}=\boxed{\dfrac{17}{21}} .

10 Helpful 0 Interesting 1 Brilliant 0 Confused
Arijit Banerjee
Mar 3, 2014

Probability for missing the target for "Robeen Hooot" is 2/3 and for " John waeen" is 2/7 . So probability of missing the target by both of them is 4/21. Now on subtracting it from 1 we get 17/21 .

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Deb Ghosal
Mar 3, 2014

Best way to solve it is to find out none of them can hit the target. So Robin can't hit in 2/3 and wayen can't hit the target in 2/7 so together can't hit the target is 2/3 * 2/7 = 4/21

So atleast one of them can hit the target is 1- 4/21 = 17/21

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Let the probability of hit by Robeen be p(A) and by John be p(B).. if hit by atleast one means p(AUB)..By using addition rule of probability,p(AUB)= p(A)+p(B)-p(A intersection B).. since A and B are independent events, p(A intersection B) is p(A) p(B).. Hence, by addition rule it becomes p(A)+p(B)-p(A) p(B) => (1/3) + (5/7) - (1/3) * (5/7) =17/21..

Vinay HN - 7 years, 2 months ago

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