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A bag contains 4 red, 5 blue and 3 green balls. If two balls are drawn at random from the bag, then which of the following statements are correct?

(A) The probability that both balls are red is 111\frac{1}{11}.

(B) The probability that one ball is red, and one ball is blue is 1033\frac{10}{33}.

(C) The probability that both balls are blue is 533\frac{5}{33}.

(D) The probability that both balls are green is 511\frac{5}{11}.

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

The bag contains:

Red balls=4\text{Red balls} = 4

Blue balls=5\text{Blue balls} = 5

Green balls=3\text{Green balls} = 3

Total balls=12\text{Total balls} = 12


When drawing 2 balls at random, we first need the total number of ways to select 2 balls from 12.

Using combinations:

(122)=12×112=66\binom{12}{2} = \frac{12 \times 11}{2} = 66

This will be our denominator for all probability calculations.


Checking Statement (A): The probability that both balls are red is 111\frac{1}{11}

Number of ways to choose 2 red balls from 4:

(42)=4×32=6\binom{4}{2} = \frac{4 \times 3}{2} = 6

Probability:

P(both red)=666=111P(\text{both red}) = \frac{6}{66} = \frac{1}{11}

Statement (A) is correct.


Checking Statement (B): The probability that one ball is red and one ball is blue is 1033\frac{10}{33}

Number of ways to choose 1 red from 4 and 1 blue from 5:

(41)×(51)=4×5=20\binom{4}{1} \times \binom{5}{1} = 4 \times 5 = 20

Probability:

P(one red, one blue)=2066=1033P(\text{one red, one blue}) = \frac{20}{66} = \frac{10}{33}

Statement (B) is correct.


Checking Statement (C): The probability that both balls are blue is 533\frac{5}{33}

Number of ways to choose 2 blue balls from 5:

(52)=5×42=10\binom{5}{2} = \frac{5 \times 4}{2} = 10

Probability:

P(both blue)=1066=533P(\text{both blue}) = \frac{10}{66} = \frac{5}{33}

Statement (C) is correct.


Checking Statement (D): The probability that both balls are green is 511\frac{5}{11}

Number of ways to choose 2 green balls from 3:

(32)=3×22=3\binom{3}{2} = \frac{3 \times 2}{2} = 3

Probability:

P(both green)=366=122P(\text{both green}) = \frac{3}{66} = \frac{1}{22}

The statement claims 511\frac{5}{11}, but the actual probability is 122\frac{1}{22}

Statement (D) is incorrect.


Statements (A), (B), and (C) are correct.

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