A dice is thrown twice. Find the probability of getting an odd number in the first throw and a multiple of 3 in the second throw.

$\frac{1}{6}$

Odd no. on the first throw = 1, 3, 5

Multiple of 3 on the second throw = 3, 6

Now let's see the combination of these two events happening as the question says = (1,3) , (1,6) , (3,3) , (3,6) , (5,6) , (5,3)

Total no. of favorable outcomes = 6

Total no. of outcomes after throwing two times of dice = 6² = 36

Required probability = \(\frac{No.\;of\;favorable\;outcomes}{Total\;no.\;of\; outcomes}\) = 6/36 = 1/6