Words from the letters of the word PROBABILITY are formed by taking all letters at a time. The probability that both B 's are not together and both I 's are not together is

Question

Words from the letters of the word PROBABILITY are formed by taking all letters at a time. The probability that both B 's are not together and both I 's are not together is (A) 52 / 55 (B) 53 / 55 (C) 54 / 55 (D) None of these

Tardigrade Admin · Accepted Answer

The total number of cases is 11 ! / 2 ! × 2 !. The number of favorable cases is [11 ! /(2 ! × 2 !)]-9 ! Therefore, the required probability is 1-(9 ! × 4/11 !)=(53/55)

Q. Words from the letters of the word PROBABILITY are formed by taking all letters at a time. The probability that both $B$ 's are not together and both $I$ 's are not together is

$52/55$

$53/55$

$54/55$

None of these

The total number of cases is $11! /2! \times 2!$ . The number of favorable cases is $[11! / (2! \times 2!)] - 9!$ Therefore, the required probability is
$1 - \frac{9 ! \times 4}{11 !} = \frac{53}{55}$

Q. Words from the letters of the word PROBABILITY are formed by taking all letters at a time. The probability that both B 's are not together and both I 's are not together is

52/55

53/55

54/55