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 D $\frac{4}{5}$
Let E be the event that the minimum of the two numbers is less than 4.
Then P(E) = 1 - P(both numbers are greater than or equal to four)
$\Rightarrow P (E) = 1 - \frac{3}{6} \times \frac{2}{5} = 1 - \frac{1}{5} = \frac{4}{5}$

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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 D 45Let E be the event that the minimum of the two numbers is less than 4. Then P(E) = 1 - P(both numbers are greater than or equal to four) ⇒P(E)=1−36×25=1−15=45

The correct option is D $\frac{4}{5}$
Let E be the event that the minimum of the two numbers is less than 4.
Then P(E) = 1 - P(both numbers are greater than or equal to four)
$\Rightarrow P (E) = 1 - \frac{3}{6} \times \frac{2}{5} = 1 - \frac{1}{5} = \frac{4}{5}$