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Probability

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Question : 97 of 127

Marks: +1, -0

Four dice are thrown, find the probability that 1 will appear on at least one of the dice.

$\frac{616}{1296}$
$\frac{308}{1296}$
$\frac{671}{1296}$
$\frac{72}{1296}$

Solution:

Concept:

Use complement rule:

P(\text{at least one 1}) = 1 - P(\text{no 1 on any die})

.

Explanation:

Total outcomes when four dice are thrown =

6^4 = 1296

.

Probability of not getting 1 on a single die =

\frac{5}{6}

.

For four dice,

P(\text{no 1}) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296}

.

Thus,

P(\text{at least one 1}) = 1 - \frac{625}{1296} = \frac{1296 - 625}{1296} = \frac{671}{1296}

.

Answer:

\frac{671}{1296}

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