Question

There are twenty cards. Ten of these cards have the letter $I$ printed on them and the other $10$ have the letter $T$ printed on them. If three cards are picked up at random and kept in the same order, the probability of making word $\text{IIT}$ is:

• A. $\frac{4}{27}$
• B. $\frac{5}{38}$
• C. $\frac{1}{8}$
• D. $\frac{9}{80}$

Answer 1

The correct option is B $\frac{5}{38}$

Consider the following events,
$A:$ getting a card with mark $\text{I}$ in first draw.
$B:$ getting a card with mark $\text{I}$ in second draw.
$C:$ getting a card with mark $\text{T}$ in this draw.
Then, the required probability is
$P(A\cap B\cap C)=P\left(A\right)P\left(\frac{B}{A}\right)P\left(\frac{C}{A\cap B}\right)=\frac{10}{20}\times \frac{9}{19}\times \frac{10}{18}=\frac{5}{38}$