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(C) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.We need to find the outcomes where the product of the two numb

Vedclass · Accepted Answer

$\frac{4}{9}$

Vedclass · Answer

(C) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.We need to find the outcomes where the product of the two numbers is less than $9$.The favorable outcomes are:For $1$ on the first die: $(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$ (Products: $1, 2, 3, 4, 5, 6$)For $2$ on the first die: $(2,1), (2,2), (2,3), (2,4)$ (Products: $2, 4, 6, 8$)For $3$ on the first die: $(3,1), (3,2)$ (Products: $3, 6$)For $4$ on the first die: $(4,1), (4,2)$ (Products: $4, 8$)For $5$ on the first die: $(5,1)$ (Product: $5$)For $6$ on the first die: $(6,1)$ (Product: $6$)Counting these,we have $6 + 4 + 2 + 2 + 1 + 1 = 16$ favorable outcomes.The probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{16}{36} = \frac{4}{9}$.

Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than $9$.

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