Two numbers are selected randomly from the set $S = {1, 2, 3, 4, 5, 6}$ without replacement one-by-one. The probability that minimum of the two numbers is less than $4$ is

\frac{1}{15}

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\frac{14}{15}

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\frac{1}{5}

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\frac{4}{5}

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Solution

The correct option is C $\frac{4}{5}$
Given, $S = {1, 2, 3, 4, 5, 6}$
Total ways $= 2! {^{6} C}_{2} = 30$
Favorable cases $30 - 6 = 24$
Therefore, the probability of the numbers less than $4$ $= \frac{24}{30} = \frac{4}{5}$ .

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Two numbers are selected randomly from the set S={1,2,3,4,5,6} without replacement one-by-one. The probability that minimum of the two numbers is less than 4 is

The correct option is C 45Given, S={1,2,3,4,5,6}Total ways =2!6C2=30Favorable cases 30−6=24Therefore, the probability of the numbers less than 4 =2430=45.

Two numbers are selected randomly from the set $S = {1, 2, 3, 4, 5, 6}$ without replacement one-by-one. The probability that minimum of the two numbers is less than $4$ is

The correct option is C $\frac{4}{5}$
Given, $S = {1, 2, 3, 4, 5, 6}$
Total ways $= 2! {^{6} C}_{2} = 30$
Favorable cases $30 - 6 = 24$
Therefore, the probability of the numbers less than $4$ $= \frac{24}{30} = \frac{4}{5}$ .