Multiple choice

A box $B_{1}$ contains $1$ white ball, $3$ red balls and $2$ black balls. Another box $B_{2}$ contains $2$ white balls, $3$ red balls and $4$ black balls. A third box $B_{3}$ contains $3$ white balls, $4$ red balls and $5$ black balls. If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$, the probability that all 3 drawn balls are of the same colour is

  1. $\displaystyle \frac{82}{648}$
  2. $\displaystyle \frac{90}{648}$
  3. $\displaystyle \frac{558}{648}$
  4. $\displaystyle \frac{566}{648}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total balls in B1=6, B2=9, B3=12. Probability of white = (1/6)(2/9)(3/12) = 6/648. Probability of red = (3/6)(3/9)(4/12) = 36/648. Probability of black = (2/6)(4/9)(5/12) = 40/648. Sum = (6+36+40)/648 = 82/648.

AI explanation

The probability of drawing balls of the same colour from the three boxes is calculated by adding the individual probabilities for each color. For white balls, the probability is (1/6) times (2/9) times (3/12), which equals 1/108. The probability for red balls is (3/6) times (3/9) times (4/12), resulting in 1/18. The probability for black balls is (2/6) times (4/9) times (5/12), which equals 5/81. Adding these probabilities gives 1/108 + 6/108 + 20/324, which totals 82/648.

Keep practicing — related questions