Multiple choice

A box contains $6$ red and $4$ white marbles. A marbels is drawn and replaced three times from the box. The probability that one white marbles is drawn

  1. $\frac{{53}}{{125}}$
  2. $\frac{{54}}{{125}}$
  3. $\frac{{56}}{{125}}$
  4. $\frac{{52}}{{125}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Probability of white (p) = 4/10 = 0.4. Probability of red (q) = 6/10 = 0.6. We want exactly one white in three draws (with replacement). This is a binomial distribution: 3C1 * (0.4)^1 * (0.6)^2 = 3 * 0.4 * 0.36 = 1.2 * 0.36 = 0.432. 0.432 = 432/1000 = 54/125.

AI explanation

Since the marble is replaced, the probability of drawing a white marble remains 4/10 or 2/5 each time, and the probability of a red marble is 3/5. We use the binomial probability formula to find the probability of exactly one white marble in three draws: 3C1 * (2/5)^1 * (3/5)^2. This calculates to 3 * (2/5) * (9/25) = 54/125.

Keep practicing — related questions