Question

Two brothers go hunting. The first brother kills an average of $3$ animals every $7$ shots and the second brother kills one animal every $3$ shots. If the two fired at the same animal at the same time, what is the probability that they will obtain a kill?

• A. $\frac{16}{21}$
• B. $\frac{15}{21}$
• C. $\frac{17}{21}$
• D. none of these

Answer 1

The correct option is A $\frac{16}{21}$

First Brother $:$ Kills average of $3$ animals on every $7$ shots

i.e, kills $3$ animals in $7$ shots.

$\therefore$ No. of shots $=3\times 7=21.$

Second Brother $:$ Kills one animal in every $3$ shots

$\Rightarrow P$(they will obtain a kill ) $=\frac{(3+1)\times 4}{21}$

$=\frac{16}{21}.$

Hence, the answer is $\frac{16}{21}.$