A bag contains 5 white, 4 black and 3 red balls. Which of the following statements are True?
A. The probability of drawing three red balls one by one without replacement is \( \displaystyle\frac{1}{220} \)
B. One by one three balls drawn without replacement then probability that third ball is red is \( \displaystyle\frac{1}{4} \)
C. The probability of drawing 2 white balls one by one without replacement is \( \displaystyle\frac{3}{4} \)
D. One by one 3 balls are drawn without replacement then probability that third ball is red is \( \displaystyle\frac{3}{4} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A and B only
The bag has 5 + 4 + 3 = 12 balls, of which 3 are red and 5 are white.
Three red balls one by one without replacement:
\( \displaystyle P=\frac{3}{12}\times\frac{2}{11}\times\frac{1}{10} \)
\( \displaystyle =\frac{6}{1320} \)
\( \displaystyle =\frac{1}{220} \)
Third ball red: before any ball is seen, each of the 12 balls is equally likely to be the third ball drawn, so the probability is the same as for the first draw:
\( \displaystyle P(\text{third ball red})=\frac{3}{12} \)
\( \displaystyle =\frac{1}{4} \)
Two white balls one by one without replacement:
\( \displaystyle P=\frac{5}{12}\times\frac{4}{11} \)
\( \displaystyle =\frac{20}{132} \)
\( \displaystyle =\frac{5}{33} \)
So P(three red) = \( \frac{1}{220} \), P(third ball red) = \( \frac{1}{4} \) and P(two white) = \( \frac{5}{33} \).
Checking the statements
(A) The probability of drawing three red balls one by one without replacement is \( \frac{1}{220} \) — Correct. \( \frac{3}{12}\times\frac{2}{11}\times\frac{1}{10}=\frac{1}{220} \).
(B) One by one three balls drawn without replacement then probability that third ball is red is \( \frac{1}{4} \) — Correct. It equals the first-draw probability \( \frac{3}{12}=\frac{1}{4} \).
(C) The probability of drawing 2 white balls one by one without replacement is \( \frac{3}{4} \) — Incorrect. The probability is \( \frac{5}{33} \).
(D) One by one 3 balls are drawn without replacement then probability that third ball is red is \( \frac{3}{4} \) — Incorrect. The probability is \( \frac{1}{4} \); \( \frac{3}{4} \) is the probability that the third ball is not red.
Tip: Without replacement, the probability that the k-th ball has a given colour, when the earlier balls are not observed, equals the probability for the first ball.