Question

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 $121\times P\left(A\right)$.

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Answer 1

Probability of drawing a red ball in the first time is $\frac{numberofredballs}{totalnumberofballs}=\frac{7}{11}$

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

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

$\therefore P\left(A\right)=\frac{7}{11}\times \frac{7}{11}=\frac{49}{121}$

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