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An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by the one without replacement. What is the probability that at least one ball is black?

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\frac{15}{12}

\frac{14}{15}

\frac{15}{18}

\frac{15}{22}

answer is D.

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Detailed Solution

Given,
Total white balls=3
Total red balls=4
Total black balls=5
Total balls,
=3+4+5
=12
Total non-black balls,
=12-5
=7
We have
A = Red or white ball in the 1st draw
B = Red or white ball in the 2nd draw
So,

P (A) = \frac{7}{12}

Also,

P (\frac{B}{A}) = \frac{6}{11}

Now we know that,

P (A \cap B) = P (A) \times P (\frac{B}{A})

P (A \cap B) = \frac{7}{12} \times \frac{6}{11}

P (A \cap B) = \frac{7}{22}

The probability that at least one ball is black,

= 1 - P (A \cap B)

= 1 - \frac{7}{22}

= \frac{15}{22}

Hence, the required probability

= \frac{15}{22}

Option 4 is correct.

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