There are 3 bags .Bag 1 contain 2 red and a 2 4 a 18 black balls, bag 2 contains 1 red and a 2 − 4 a + 9 black balls and bag 3 contains 3 red and a 2 − 4 a + 7 black balls .A ball is draw at random fr

Question

There are 3 bags .Bag 1 contain 2 red and $a^{2} 4 a 18$ black balls, bag 2 contains 1 red and $a^{2} - 4 a + 9$ black balls and bag 3 contains 3 red and $a^{2} - 4 a + 7$ black balls .A ball is draw at random from at random chosen bag. Then maximum value of probability that it is a red ball is

Options:

Answer:

A

Solution:

Req. prob $= \frac{1}{3} (\frac{6}{a ^{2} - 4 a + 1 c})$

$(P (A))_{m a x} = \frac{1}{3} \times \frac{6}{6} = \frac{1}{3}$

TestHub · Accepted Answer

Req. prob = 3 1 ​ ( a 2 − 4 a + 1 c 6 ​ ) ( P ( A ) ) m a x ​ = 3 1 ​ × 6 6 ​ = 3 1 ​

Mathematics - Probability Question with Solution | TestHub

Options:

Doubts & Discussion