question_answer

For 6 trials of an experiment, let X be a binomial variate which satisfies the relation 9P(X = 4) = P(X = 2). Find the probability of success.

OR

A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 balls. A die is thrown. If 1 or 2 appears on it, then bag A is chosen, otherwise bag B. If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black.

Answer:

Let p = Probability of success

and q = Probability of failure

Then,\[9\,P\,(X=4)=P\,(X=2)\]

\[9\,\,{}^{6}{{C}_{4}}{{p}^{4}}{{q}^{2}}={}^{6}{{C}_{2}}{{p}^{2}}{{q}^{4}}\Rightarrow 9{{p}^{2}}={{q}^{2}}\]

\[\therefore \]                  \[q=3\,p\]

Also,     \[p+q=1\Rightarrow -+3p=1\Rightarrow p=\frac{1}{4}\]

OR

Probability of getting 1 or 2 or die\[={{p}_{1}}=\frac{2}{6}=\frac{1}{3}.\]

Probability of not getting 1 or 2\[={{p}_{2}}=1-\frac{1}{3}=\frac{2}{3}.\]

Bag A (4 black balls and 6 red)\[\to \]10 balls.

P (a red and a black ball)\[={{P}_{A}}=\frac{{}^{4}{{C}_{1}}\times {}^{6}{{C}_{1}}}{{}^{10}{{C}_{2}}}\]

\[=\frac{2\times 4\times 6}{10\times 9}=\frac{8}{15}\]

Bag B (7 black balls and 3 red) \[\to \]10 balls

P (a red ball and a black ball)\[={{P}_{B}}=\frac{{}^{7}{{C}_{1}}\times {}^{3}{{C}_{1}}}{{}^{10}{{C}_{2}}}\]

\[=\frac{2\times 7\times 3}{10\times 9}=\frac{7}{15}\]

Hence, required probability

= (1 or 2 occurs AND balls are selected from bag A) or (1 or 2 doesn?t occurs AND balls are selected from bag B)

\[={{P}_{1}}\times {{P}_{A}}+{{P}_{2}}\times {{P}_{B}}\]

\[=\frac{1}{3}\times \frac{8}{15}+\frac{2}{3}\times \frac{7}{15}\]

\[=\frac{8}{45}+\frac{14}{45}=\frac{22}{45}\]