A bag contains 13 red,14 green,and 15 black b | vedclass

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(B) For $P_1$,the total number of balls is $13 + 14 + 15 = 42$. The number of ways to choose $4$ balls out of $42$ is $^{42}C_4$. The number of ways t

Vedclass · Accepted Answer

$P_1 > P_2$

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(B) For $P_1$,the total number of balls is $13 + 14 + 15 = 42$. The number of ways to choose $4$ balls out of $42$ is $^{42}C_4$. The number of ways to choose $2$ black balls from $15$ and $2$ non-black balls from $27$ is $^{15}C_2 	imes ^{27}C_2$.$P_1 = \frac{^{15}C_2 	imes ^{27}C_2}{^{42}C_4} = \frac{105 	imes 351}{111930} = \frac{36855}{111930} \approx 0.329$.For $P_2$,the number of balls of each color is doubled,so there are $26$ red,$28$ green,and $30$ black balls. Total balls = $84$. We choose $8$ balls. The probability of getting exactly $4$ black balls is $P_2 = \frac{^{30}C_4 	imes ^{54}C_4}{^{84}C_8}$.Calculating this ratio,we find $P_2 \approx 0.274$.Comparing the values,$P_1 > P_2$.

$A$ bag contains $13$ red,$14$ green,and $15$ black balls. The probability of getting exactly $2$ black balls when pulling out $4$ balls is $P_1$. Now,the number of balls of each color is doubled,and $8$ balls are pulled out. The probability of getting exactly $4$ black balls is $P_2$. Then:

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