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 IIT is:
A. $\dfrac{4}{{27}}$
B. $\dfrac{5}{{38}}$
C. $\dfrac{1}{8}$
D. $\dfrac{9}{{80}}$

Answer 1

Hint: To calculate the probability, of picking three cards at random and kept in the same order, the probability of making word IIT will be the intersection of getting first card with “I” printed on it with getting second card with “I” printed on it and getting third card with “T” printed on it. The probability is calculated using $P\left( {A \cap B \cap C} \right) = P\left( A \right) \times P\left( {\dfrac{B}{A}} \right) \times P\left( {\dfrac{C}{{A \cap B}}} \right)$

Complete step by step answer
We have to select three cards.
Let $A$ be an event of getting the first card with “I” printed on it.
Let $B$ be an event of getting a second card with “I” printed on it.
Let $C$ be an event of getting a third card with “T” printed on it.
We have to find the probability of getting the first card with “I” printed on it, the second card with “I” printed on it and getting the third card with “T” printed on it.
This implies we have to find the intersection of $A$, $B$ and $C$.
That is we will calculate $P\left( {A \cap B \cap C} \right)$.
The formula for calculating is the product of probability of event $A$ with probability of $B$when event $A$ has already taken place and probability of $C$ when both events $A$ and $B$ have taken place.
That is, $P\left( {A \cap B \cap C} \right) = P\left( A \right) \times P\left( {\dfrac{B}{A}} \right) \times P\left( {\dfrac{C}{{A \cap B}}} \right)$
Determine the probability of event $A$.
There are a total of 20 cards and out of them, ten have “I” printed on it.
Thus, the number of favourable outcomes is 10 and total number of outcomes are 20.
Probability of an event is $\dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of outcomes}}}}$
Therefore, $P\left( A \right) = \dfrac{{10}}{{20}}$
Next, calculate the probability of $B$ when event $A$ has already taken place.
Since, one card is already drawn, there are now 19 cards and “I” is printed on 9 of them as one card with “I” printed on it is selected in the event $A$.
Therefore, number of favourable outcomes is 9 and total number of outcomes are 19
And, $P\left( {\dfrac{B}{A}} \right) = \dfrac{9}{{19}}$
Similarly, calculate the probability of $C$ when both events $A$ and $B$ have taken place.
Now, there are 18 cards left as two cards are already drawn in event $A$ and $B$. From the left cards, “T” is printed on 10 of them.
Therefore, number of favourable outcomes is 10 and total number of outcomes are 18
And, $P\left( {\dfrac{C}{{A \cap B}}} \right) = \dfrac{{10}}{{18}}$
Hence,
 $
  P\left( {A \cap B \cap C} \right) = P\left( A \right) \times P\left( {\dfrac{B}{A}} \right) \times P\left( {\dfrac{C}{{A \cap B}}} \right) \\
  P\left( {A \cap B \cap C} \right) = \dfrac{{10}}{{20}} \times \dfrac{9}{{19}} \times \dfrac{{10}}{{18}} \\
  P\left( {A \cap B \cap C} \right) = \dfrac{{900}}{{6,840}} \\
  P\left( {A \cap B \cap C} \right) = \dfrac{5}{{38}} \\
$
Hence, option B is correct.

Note: The probability of an event cannot be less than 0 and be greater than 1. Probability for an event is calculated as $\dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of outcomes}}}}$. In this question, be careful while finding probability of $P\left( {\dfrac{B}{A}} \right)$ and $P\left( {\dfrac{C}{{A \cap B}}} \right)$ as the number of favourable outcomes will depend on the cards left.