Question

The letters of the word $\text{PROBABILITY}$ are written down at random in a row. Let $E_1$ denote the event that two $I^{\prime }s$ are together and $E_2$ denote the event that two $B^{\prime }s$ are together. Then which of the following is (are) CORRECT?

• A. $P\left(E_1\right)=P\left(E_2\right)=\frac{2}{11}$
• B. $P(E_1\cap E_2)=\frac{2}{55}$
• C. $P(E_1\cup E_2)=\frac{18}{55}$
• D. $P\left(E_1|E_2\right)=\frac{1}{5}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $P\left(E_1\right)=P\left(E_2\right)=\frac{2}{11}$
B $P(E_1\cap E_2)=\frac{2}{55}$
C $P(E_1\cup E_2)=\frac{18}{55}$
D $P\left(E_1|E_2\right)=\frac{1}{5}$

$PROBABILITY$
$B\to 2,I\to 2,P\to 1,R\to 1,O\to 1,A\to 1,$
$L\to 1,T\to 1,Y\to 1$

$P\left(E_1\right)=P\left(E_2\right)=\frac{\frac{10!}{2!}}{\frac{11!}{2!\cdot 2!}}=\frac{2}{11}$

$P(E_1\cap E_2)=\frac{9!}{\frac{11!}{2!\cdot 2!}}=\frac{2}{55}$

$P(E_1\cup E_2)=P\left(E_1\right)+P\left(E_2\right)-P(E_1\cap E_2)=\frac{2}{11}+\frac{2}{11}-\frac{2}{55}=\frac{18}{55}$

$P\left(E_1|E_2\right)=\frac{P(E_1\cap E_2)}{P\left(E_2\right)}=\frac{\frac{2}{55}}{\frac{2}{11}}=\frac{1}{5}$