Question

A bag contains 20 green marbles and 12 blue marbles. Four marbles are drawn at random. What is the probability of getting all green marbles?

• A. $\text{P(E)}=\frac{95}{248}\times \frac{51}{145}$
• B. $\text{P(E)}=\frac{323}{248}\times \frac{51}{145}$
• C. $\text{P(E)}=\frac{15}{248}\times \frac{95}{145}$

Answer 1

The correct option is A $\text{P(E)}=\frac{95}{248}\times \frac{51}{145}$

Given that,
Total marbles $=20+12=32$
Number of Green marbles $=20$
Number of Blue marbles $=12$

Probability of choosing first green marble is $\text{P(A)}=\frac{20}{32}=\frac{5}{8}$

Probability of choosing second green marble is $\text{P(B)}=\frac{19}{31}$

Probability of choosing third green marble is $\text{P(C)}=\frac{18}{30}=\frac{3}{5}$

Probability of choosing fourth green marble is $\text{P(D)}=\frac{17}{29}$

We have to find the probability of getting all four marbles of green color.

So, the probability of getting all four green marbles is $\text{P(A and B and C and D)}=\text{P(E)}=\text{P(A)}\times \text{P(B)}\times \text{P(C)}\times \text{P(D)}$

$\Rightarrow \text{P(E)}=\frac{5}{8}\times \frac{19}{31}\times \frac{3}{5}\times \frac{17}{29}$

$\Rightarrow \text{P(E)}=\frac{95}{248}\times \frac{51}{145}$