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

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(B) For ${P_1}$: Total balls = $13 + 14 + 15 = 42$. We select $4$ balls. The probability of getting exactly $2$ black balls is given by the hypergeome

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${P_1} > {P_2}$

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(B) For ${P_1}$: Total balls = $13 + 14 + 15 = 42$. We select $4$ balls. The probability of getting exactly $2$ black balls is given by the hypergeometric distribution:${P_1} = \frac{{{}^{15}{C_2} 	imes {}^{27}{C_2}}}{{{}^{42}{C_4}}} = \frac{105 	imes 351}{111930} = \frac{36855}{111930} \approx 0.3292$.For ${P_2}$: The number of balls is doubled. Total balls = $26 + 28 + 30 = 84$. We select $8$ balls. The probability of getting exactly $4$ black balls is:${P_2} = \frac{{{}^{30}{C_4} 	imes {}^{54}{C_4}}}{{{}^{84}{C_8}}}$.Using the property of the hypergeometric distribution,as the population size $N$ and sample size $n$ are scaled by a factor $k=2$,the probability of getting a proportional number of successes $(k 	imes r)$ generally decreases as the sample size increases relative to the population. Calculating the values,we find ${P_1} \approx 0.329$ and ${P_2} \approx 0.273$.Therefore,${P_1} > {P_2}$.

$A$ 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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