Question

The probability of a boy getting a scholarship is 0.90 and that of a girl student getting a scholarship is 0.80. The probability that both getting scholarship is:
A. $\dfrac{{98}}{{100}}$
B. $\dfrac{2}{{100}}$
C. $\dfrac{{72}}{{100}}$
D. $\dfrac{{28}}{{100}}$

Answer 1

Hint: Given that, the probability of a boy student getting scholarship \[P\left( B \right) = 0.90\] and the probability of a girl student getting a scholarship \[P\left( G \right) = 0.80\] .
It is asked to find the probability that both boy student and girl student get a scholarship.
For that, we have to find $P\left( {B \cap G} \right)$ .
Here, if two events are independent, then the probability $P\left( {B \cap G} \right)$ is the product of both individual probabilities.

Complete step-by-step answer:
Let the event that a boy student gets a scholarship be B and a girl student getting a scholarship be G.
Thus, the probability of a boy student getting scholarship \[P\left( B \right) = 0.90\] and the probability of a girl student getting a scholarship \[P\left( G \right) = 0.80\] .
We are asked to find the probability that both boy student and girl student get a scholarship i.e. $P\left( {B \cap G} \right)$ .
Now, we have to find $P\left( {B \cap G} \right)$ .
It is clear from the question that, the given two events are independent from each other.
We know, for independent events, the probability of a boy and a girl getting a scholarship is the product of individual probabilities i.e. product of probability that a boy student gets a scholarship and probability that a girl student gets a scholarship.
 $\therefore P\left( {B \cap G} \right) = P\left( B \right) \times P\left( G \right)$
                \[ = \left( {0.90} \right)\left( {0.80} \right)\]
               = 0.72
Thus, the probability that both a boy student and girl student get a scholarship is 0.72 $ = \dfrac{{72}}{{100}}$ .

Note: Here, we have to take note that the given two events are independent and not, mutually exclusive events.
Independent events:
If any event B is not being affected by the previous event A, then both the event A and B are called independent events.
Thus, $P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)$ .
For example, find the probability of getting 3 heads on tossing a coin three times. Here, P(Head) = 0.5. Now, the second head would not be affected by the first head. So, the probability of three heads will be \[0.5 \times 0.5 \times 0.5 = 0.125\] .
Mutually Exclusive events:
If any two events, say A and B, are mutually exclusive, it is not possible that they both occur together.
Thus, $P\left( {A \cap B} \right) = 0$ .
For example, out of 52 cards one card is chosen, then find the probability that the chosen card is a queen card and a king card. Here, one card cannot be both a queen card and a king card. So, the probability of it happening becomes 0.