A box contains 10 black and 5 red balls. Two balls are drawn from the box one after the other without replacement, then probability that
(A) Both balls are black = 3/7
(B) Both balls are red = 5/21
(C) First is black and second is red = 2/21
(D) First is red and second is black = 5/21
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (D) only
The box has 15 balls. Without replacement, the second ball is drawn from the remaining 14 balls, so by the multiplication rule \( P(E\cap F)=P(E)\,P(F|E) \):
\( \displaystyle P(\text{both black})=\frac{10}{15}\times\frac{9}{14}=\frac{3}{7} \)
\( \displaystyle P(\text{both red})=\frac{5}{15}\times\frac{4}{14}=\frac{2}{21} \)
\( \displaystyle P(\text{black, then red})=\frac{10}{15}\times\frac{5}{14}=\frac{5}{21} \)
\( \displaystyle P(\text{red, then black})=\frac{5}{15}\times\frac{10}{14}=\frac{5}{21} \)
So P(both black) = 3/7, P(both red) = 2/21, P(black, then red) = 5/21 and P(red, then black) = 5/21.
Checking the statements
(A) Both balls are black = 3/7 — Correct. \( \frac{10}{15}\times\frac{9}{14}=\frac{3}{7} \).
(B) Both balls are red = 5/21 — Incorrect. The probability is \( \frac{5}{15}\times\frac{4}{14}=\frac{2}{21} \).
(C) First is black and second is red = 2/21 — Incorrect. The probability is \( \frac{10}{15}\times\frac{5}{14}=\frac{5}{21} \).
(D) First is red and second is black = 5/21 — Correct. \( \frac{5}{15}\times\frac{10}{14}=\frac{5}{21} \).