A bag has 13 red, 14 green and 15 black balls | vedclass

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Question

(b) ${P_1} = \frac{{{}^{15}{C_2} 	imes {}^{27}{C_2}}}{{{}^{42}{C_4}}} = \frac{{27}}{{82}}$

and ${P_2} = \frac{{{}^{30}{C_4} 	imes {}^{54}{C_4}}}{

Vedclass · Accepted Answer

${P_1} > {P_2}$

Vedclass · Answer

(b) ${P_1} = \frac{{{}^{15}{C_2} 	imes {}^{27}{C_2}}}{{{}^{42}{C_4}}} = \frac{{27}}{{82}}$

and ${P_2} = \frac{{{}^{30}{C_4} 	imes {}^{54}{C_4}}}{{{}^{84}{C_8}}} = \frac{{17\,.\,29\,.\,45\,.\,53}}{{11\,.\,79\,.\,82\,.\,83}}$     (After simplification)

Hence ${P_1} > {P_2}.$

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

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