The probability of getting a sum of 3,5,or 11 | vedclass

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(C) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.Favorable outcomes for a sum of $3$: $(1, 2), (2, 1)$ (Total $

Vedclass · Accepted Answer

$\frac{2}{9}$

Vedclass · Answer

(C) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.Favorable outcomes for a sum of $3$: $(1, 2), (2, 1)$ (Total $2$ cases).Favorable outcomes for a sum of $5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ (Total $4$ cases).Favorable outcomes for a sum of $11$: $(5, 6), (6, 5)$ (Total $2$ cases).Total favorable outcomes $= 2 + 4 + 2 = 8$.Required probability $= \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{8}{36} = \frac{2}{9}$.

The probability of getting a sum of $3$,$5$,or $11$ when throwing two dice is:

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