A Box B1 contains 3 blue balls and 6 red balls Another Box B2 contains 8 blue balls and 'n' red balls (n ∈ N) . A ball selected at random from a box is found to be red. If p is the probability that this red ball drawn is from box B2, then

Question

A Box B1 contains 3 blue balls and 6 red balls Another Box B2 contains 8 blue balls and 'n' red balls (n ∈ N) . A ball selected at random from a box is found to be red. If p is the probability that this red ball drawn is from box B2, then (A) (1/7) ≤ p< (3/5) (B) (3/5) ≤ p< 1 (C) 0< p ≤ (3/5) (D) 0 ≤ p ≤ (1/7)

Tardigrade Admin · Accepted Answer

According to given information, the required probability

= \frac{\frac{1}{2} \times \frac{n}{n + 8}}{( \frac{1}{2} \times \frac{n}{n + 8} ) + ( \frac{1}{2} \times \frac{6}{3 + 6} )}

(According to Baye's theorem)

= \frac{n + 8}{\frac{n}{n + 8} + \frac{2}{3}} = \frac{3 n}{3 n + 2 n + 16} = \frac{3 n}{5 n + 16} = p

(given)
For minimum value of

p, n

must be 1 (as

n \in N)

So,

p \geq \frac{3}{21}

\Rightarrow p \geq \frac{1}{7}

For maximum value of

p, n \to \infty

So,

p < \frac{3}{5}

\Rightarrow \frac{1}{7} \leq p < \frac{3}{5}

Q. A Box $B_{1}$ contains $3$ blue balls and $6$ red balls Another Box $B_{2}$ contains $8$ blue balls and 'n' red balls $(n \in N) .$ A ball selected at random from a box is found to be red. If $p$ is the probability that this red ball drawn is from box $B_{2}$ , then

$\frac{1}{7} \leq p < \frac{3}{5}$

$\frac{3}{5} \leq p < 1$

$0 < p \leq \frac{3}{5}$

$0 \leq p \leq \frac{1}{7}$

Q. A Box B1​ contains 3 blue balls and 6 red balls Another Box B2​ contains 8 blue balls and 'n' red balls (n∈N). A ball selected at random from a box is found to be red. If p is the probability that this red ball drawn is from box B2​, then

71​≤p<53​

53​≤p<1

0<p≤53​

0≤p≤71​

Q. A Box $B_{1}$ contains $3$ blue balls and $6$ red balls Another Box $B_{2}$ contains $8$ blue balls and 'n' red balls $(n \in N) .$ A ball selected at random from a box is found to be red. If $p$ is the probability that this red ball drawn is from box $B_{2}$ , then

$\frac{1}{7} \leq p < \frac{3}{5}$

$\frac{3}{5} \leq p < 1$

$0 < p \leq \frac{3}{5}$

$0 \leq p \leq \frac{1}{7}$