Question

A die is thrown twice. What is the probability that
$\left(i\right)$ 3 will not come up either time?
$\left(ii\right)$ 6 will come up at least once?

• A. $\left(i\right)\frac{12}{25}\left(ii\right)\frac{16}{25}$
• B. $\left(i\right)\frac{18}{25}\left(ii\right)\frac{9}{25}$
• C. $\left(i\right)\frac{25}{36}\left(ii\right)\frac{11}{36}$
• D. $\left(i\right)\frac{23}{36}\left(ii\right)\frac{17}{36}$

Answer 1

Answer: (C)

$\left(i\right)\frac{25}{36}\left(ii\right)\frac{11}{36}$

When a dice is thrown twice, the possible 36 outcomes.

$3$ will not come up either time, when pair of outcomes are anything except $(1,3),(3,1),(3,3),(2,3),(3,2),(3,6),(6,3)(4,3),(3,4),(3,5),(5,3)$

Probability that $3$ will not come up either time $=\frac{36-11}{36}=\frac{25}{36}$

$6$ will come up at least once, when pair of outcomes are $(1,6),(6,1),(6,2),(2,6),(3,6),(6,3),(4,6),(6,4),(6,5),(5,6),(6,6)$

Probability that $6$ will come up at least once $=\frac{11}{36}$