The chance of throwing at least 9 in a single | vedclass

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(B) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.We need to find the probability of getting a sum of at least $

Vedclass · Accepted Answer

$\frac{5}{18}$

Vedclass · Answer

(B) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.We need to find the probability of getting a sum of at least $9$,which means the sum can be $9, 10, 11,$ or $12$.The favorable outcomes are:- Sum $= 9$: $(3,6), (4,5), (5,4), (6,3)$ ($4$ outcomes)- Sum $= 10$: $(4,6), (5,5), (6,4)$ ($3$ outcomes)- Sum $= 11$: $(5,6), (6,5)$ ($2$ outcomes)- Sum $= 12$: $(6,6)$ ($1$ outcome)Total favorable outcomes $= 4 + 3 + 2 + 1 = 10$.The required probability $= \frac{	ext{Total favorable outcomes}}{	ext{Total possible outcomes}} = \frac{10}{36} = \frac{5}{18}$.

The chance of throwing at least $9$ in a single throw with two dice is:

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