Question

In a class test, 8 out of 30 students got full marks. If three students are chosen at random, what is the probability that all three got full marks?

• A. $\text{P(E)}=\frac{4}{15}\times \frac{7}{29}\times \frac{3}{14}$
  
• B. $\text{P(E)}={\left(\frac{4}{15}\right)}^3$
  
• C. $\text{P(E)}=\frac{8}{30}\times \frac{7}{29}\times \frac{7}{28}$
  
• D. (a) and (b) above.

Answer 1

The correct option is A $\text{P(E)}=\frac{4}{15}\times \frac{7}{29}\times \frac{3}{14}$


Given,
Number of students in the class $=30$
Number of students got full marks $=8$

Let A, B, and C are events of choosing first, second, and third student who got full marks respectively.

Probability of choosing first student who got full marks is $\text{P(A)}=\frac{8}{30}=\frac{4}{15}$

Probability of choosing second student who got full marks is $\text{P(B)}=\frac{7}{29}$

Probability of choosing third student who got full marks is $\text{P(C)}=\frac{6}{28}=\frac{3}{14}$

We have to find the probablity of choosing three students without replacing.
$\Rightarrow \text{P(A,B and C)}=\text{P(E)}=\text{P(A)}\times \text{P(B)}\times \text{P(C)}$

$\Rightarrow \text{P(E)}=\frac{4}{15}\times \frac{7}{29}\times \frac{3}{14}$