If a dice is thrown twice, then find the probability of getting 1 in the first throw only.

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\frac{11}{36}

\frac{7}{36}

\frac{5}{36}

None of these

answer is C.

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Detailed Solution

Given a dice,
Probability of getting 1 in first throw,

P (M) = \frac{1}{6}

Probability of not getting 1 in second throw,

P (N) = \frac{5}{6}

Their independent events = M and N
Then,
The required probability,

P (MN) = P (M) \times P (N)

P (MN) = \frac{1}{6} \times \frac{5}{6}

P (MN) = \frac{5}{36}

Hence, the required probability is

\frac{5}{36}

.
Option 3 is correct.

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