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Miscellaneous Exercise · Q1

Q.A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that

(i) all will be blue?
(ii) atleast one will be green?
Odisha ChseTextbookSubjective· 3mImportance★★★★★
44% · 41/93 Questions
✓ Free question

Draw 55 marbles from 6060 (order irrelevant), so the total number of ways is (605)\binom{60}{5}. (i) All blue: (205)(605)\dfrac{\binom{20}{5}}{\binom{60}{5}}.

(ii) At least one green =1−(305)(605)=1-\dfrac{\binom{30}{5}}{\binom{60}{5}}, using non-green =10+20=30=10+20=30.

The box has 1010 red, 2020 blue and 3030 green marbles, a total of 10+20+30=6010+20+30=60 marbles. We draw 55 of them, and the order in which they come out does not matter — that is the signal to count with combinations. Since every group of 55 marbles is equally likely,

P(event)=number of favourable groups of 5total number of groups of 5,total=(605).P(\text{event})=\frac{\text{number of favourable groups of }5}{\text{total number of groups of }5},\qquad \text{total}=\binom{60}{5}.

(i) All five are blue

There are 2020 blue marbles, and we need all 55 drawn marbles to come from them. The number of favourable groups is (205)\binom{20}{5}, so

P(all blue)=(205)(605).P(\text{all blue})=\frac{\binom{20}{5}}{\binom{60}{5}}.

(ii) At least one is green

"At least one green" covers many cases (exactly 1,2,3,41,2,3,4 or 55 green), so it is quicker to use the complement: first find the probability of drawing no green marble, then subtract from 11.

If no marble is green, all 55 come from the non-green marbles. The non-green marbles are the reds and blues: 10+20=3010+20=30. The number of ways to choose 55 from these 3030 is (305)\binom{30}{5}, so

P(no green)=(305)(605),P(at least one green)=1−(305)(605).P(\text{no green})=\frac{\binom{30}{5}}{\binom{60}{5}},\qquad P(\text{at least one green})=1-\frac{\binom{30}{5}}{\binom{60}{5}}.

Watch out

The non-green count is 10+20=3010+20=30, not 4040. Add the red and blue marbles carefully before writing the combination.

✓Final answer

  1. (205)(605)\displaystyle \frac{\binom{20}{5}}{\binom{60}{5}}; ;
  2. 1−(305)(605)\displaystyle 1-\frac{\binom{30}{5}}{\binom{60}{5}}.

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