Multiple choice

$(A)$ A box contains $3$ blue, $2$ white and $4$ red marbles. If a marble is drawn from the box. What is the probability that it will be $(i)$ White $\quad (ii)$ Blue $\quad (iii)$ Red? $(B)$ A bag contains $5$ red, $8$ green and $7$ white balls. One ball is drawn at random from the bag, find the probability of getting $(i)$ A white ball or a green ball $(ii)$ Neither a green ball nor a red ball.

  1. $(A)\space (i)\quad \displaystyle\frac{1}{9} \quad (ii)\quad \displaystyle\frac{2}{3} \quad (iii)\quad \displaystyle\frac{5}{9} \\ (B)\space (i)\quad \displaystyle\frac{6}{13}\quad (ii)\quad \displaystyle\frac{11}{20}$
  2. $(A)\space (i)\quad \displaystyle\frac{5}{9} \quad (ii)\quad \displaystyle\frac{1}{4} \quad (iii)\quad \displaystyle\frac{2}{11} \\ (B)\space (i)\quad \displaystyle\frac{1}{2}\quad (ii)\quad \displaystyle\frac{5}{19}$
  3. $(A)\space (i)\quad \displaystyle\frac{2}{9} \quad (ii)\quad \displaystyle\frac{1}{3} \quad (iii)\quad \displaystyle\frac{4}{9} \\ (B)\space (i)\quad \displaystyle\frac{3}{4}\quad (ii)\quad \displaystyle\frac{7}{20}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For (A), total marbles = 3+2+4 = 9. P(White) = 2/9, P(Blue) = 3/9 = 1/3, P(Red) = 4/9. For (B), total balls = 5+8+7 = 20. P(White or Green) = (7+8)/20 = 15/20 = 3/4. P(Neither Green nor Red) = P(White) = 7/20.

AI explanation

For box A, the total number of marbles is 3 + 2 + 4 = 9. Using the classical definition of probability, the chance of drawing a white marble is 2/9, a blue marble is 3/9 or 1/3, and a red marble is 4/9. For bag B, the total number of balls is 5 + 8 + 7 = 20. The probability of drawing a white or green ball is (7 + 8)/20 = 15/20 = 3/4, and the probability of neither green nor red (meaning white) is 7/20. These calculations match the values given in the correct option.

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