Question

If a dice is thrown twice, then find the probability of getting 1 in the first throw only
A. $\dfrac{{11}}{{36}}$
B. $\dfrac{7}{{36}}$
C. $\dfrac{5}{{36}}$
D. None of these

Answer 1

Hint: Here we have to 1st find the probability of getting 1 in the first throw next the second throw then the condition obtained is taken as a product of both in fraction form.

Complete step-by-step answer:
Probability is given by P(A)=$\dfrac{{No{\text{ of favourable outcomes}}}}{{Total{\text{ no of outcomes }}}}$
Total no of outcomes possible when a dice is thrown up=6
Probabilities of getting 1 in first throw, $P\left( A \right) = \dfrac{1}{6}$
Probabilities of not getting 1 in second throw, $P(B) = 1 - \dfrac{1}{6} = \dfrac{5}{6}$
Both are independent events, so the required probability is
$P\left( {AB} \right) = P\left( A \right) \times P\left( B \right) = \dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{5}{{36}}$
So the correct answer is option (c).

Note: In such a problem it is evidently the same whether a single die be thrown twice or two dice be thrown at once. Also note that these two events are not related to each other, that is they are independent. So, we have multiplied these two events.