The probability that police inspector Mark will catch the thief in a day is (1/4) and the probability that he will catch a robber in a day is (1/5) and the probability that he will catch both a thief and robber in a day is (1/15) then the probability that He will catch at least 1 mischief is ?

Question

The probability that police inspector Mark will catch the thief in a day is (1/4) and the probability that he will catch a robber in a day is (1/5) and the probability that he will catch both a thief and robber in a day is (1/15) then the probability that He will catch at least 1 mischief is ? (A) (23/60) (B) (27/60) (C) (31/60) (D) (19/60)

Tardigrade Admin · Accepted Answer

Required probability =(1/4)+(1/5)-(1/15)=(15+12-4/60)=(23/60)

$\frac{23}{60}$

$\frac{27}{60}$

$\frac{31}{60}$

$\frac{19}{60}$

Required probability $= \frac{1}{4} + \frac{1}{5} - \frac{1}{15} = \frac{15 + 12 - 4}{60} = \frac{23}{60}$

Q. The probability that police inspector Mark will catch the thief in a day is 41​ and the probability that he will catch a robber in a day is 51​ and the probability that he will catch both a thief and robber in a day is 151​ then the probability that He will catch at least 1 mischief is ?

6023​

6027​

6031​

6019​

Required probability =41​+51​−151​=6015+12−4​=6023​

$\frac{23}{60}$

$\frac{27}{60}$

$\frac{31}{60}$

$\frac{19}{60}$

Required probability $= \frac{1}{4} + \frac{1}{5} - \frac{1}{15} = \frac{15 + 12 - 4}{60} = \frac{23}{60}$