MHT CET · Maths · Probability

An urn contains 4 red and 5 white balls. Two balls are drawn one after the other
without replacement, then the probability that both the balls are red is

A \(\frac{5}{6}\)
B \(\frac{1}{6}\)
C \(\frac{2}{9}\)
D \(\frac{4}{9}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{6}\)

Step-by-step Solution

Detailed explanation

Red balls \(=4\) and White balls \(=5 \Rightarrow\) Total balls \(=4+5=9\) Two balls are drawn one after the other without replacement
\(n(S)={ }^{9} C_{1} \times{ }^{8} C_{1}=9 \times 8\)
\(\text { Required probability }=\frac{{ }^{4} C_{1} \times{ }^{3} C_{1}}{9 \times 8}=\frac{4 \times 3}{9 \times 8}=\frac{1}{3 \times 2}=\frac{1}{6}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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An urn contains 4 red and 5 white balls. Two balls are drawn one after the otherwithout replacement, then the probability that both the balls are red is

Step-by-step Solution

An urn contains 4 red and 5 white balls. Two balls are drawn one after the other
without replacement, then the probability that both the balls are red is