A bag has 13 red, 14 green and 15 black balls. probability of getting exactly 2 blacks on pulling ou

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14.Probability

hard

A bag has $13$ red, $14$ green and $15$ black balls. The probability of getting exactly $2$ blacks on pulling out $4$ balls is ${P_1}$. Now the number of each colour ball is doubled and $8$ balls are pulled out. The probability of getting exactly $4$ blacks is ${P_2}.$ Then

${P_1} = {P_2}$

${P_1} > {P_2}$

${P_1} < {P_2}$

None of these

Solution

(b) ${P_1} = \frac{{{}^{15}{C_2} \times {}^{27}{C_2}}}{{{}^{42}{C_4}}} = \frac{{27}}{{82}}$

and ${P_2} = \frac{{{}^{30}{C_4} \times {}^{54}{C_4}}}{{{}^{84}{C_8}}} = \frac{{17\,.\,29\,.\,45\,.\,53}}{{11\,.\,79\,.\,82\,.\,83}}$ (After simplification)

Hence ${P_1} > {P_2}.$

Standard 11

Mathematics

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