An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. If the probability of getting 2 red balls is P(A), find $121 \times P (A)$ .
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Solution

Probability of drawing a red ball in the first time is $\frac{n u m b e r o f r e d b a l l s}{t o t a l n u m b e r o f b a l l s} = \frac{7}{11}$

After replacement, number of red balls =7 and number of blue balls remains 4.

Probability of drawing a red ball in the second chance = $\frac{7}{11}$

$∴ P (A) = \frac{7}{11} \times \frac{7}{11} = \frac{49}{121}$

$\Rightarrow$ Answer = $P (A) \times 121 = 49$

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An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. If the probability of getting 2 red balls is P(A), find 121×P(A). ___

Probability of drawing a red ball in the first time is number of red ballstotal number of balls=711 After replacement, number of red balls =7 and number of blue balls remains 4. Probability of drawing a red ball in the second chance =711 ∴P(A)=711×711=49121 ⇒ Answer = P(A)×121=49

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