A bag contains 6 red, 4 blue and 2 yellow balls. Three balls are drawn one by one with replacement. Find the probability of getting exactly one red ball.

Question

A bag contains 6 red, 4 blue and 2 yellow balls. Three balls are drawn one by one with replacement. Find the probability of getting exactly one red ball. (A) (1/4) (B) (3/8) (C) (3/4) (D) (1/2)

Tardigrade Admin · Accepted Answer

Let A be the event that ball drawn is red, then p= (6/12) = (1/2), q=1-(1/2)=(1/2) and n = 3 ∴ Probability of drawing exactly 1 red ball in 3 trials = 3C1pq2 = 3C1((1/2))((1/2))2 = (3/8).

Q. A bag contains $6$ red, $4$ blue and $2$ yellow balls. Three balls are drawn one by one with replacement. Find the probability of getting exactly one red ball.

$\frac{1}{4}$

$\frac{3}{8}$

$\frac{3}{4}$

$\frac{1}{2}$

Let $A$ be the event that ball drawn is red, then
$p = \frac{6}{12} = \frac{1}{2}$ ,
$q = 1 - \frac{1}{2} = \frac{1}{2}$ and $n = 3$
$∴$ Probability of drawing exactly $1$ red ball in $3$ trials
$=^{3} C_{1} p q^{2} =^{3} C_{1} (\frac{1}{2}) (\frac{1}{2})^{2} = \frac{3}{8}$ .

Q. A bag contains 6 red, 4 blue and 2 yellow balls. Three balls are drawn one by one with replacement. Find the probability of getting exactly one red ball.

41​

83​

43​

21​