In a single throw of three dice, find the probability of getting

Solution:

Total no.of cases will be 6 x 6 x 6 = 216(because each die can have values from 1 to 6)

Desired outcomes are those whose sum up to $5 .$ Desired outcomes are $(1,1,3),(1,3,$, $1),(1,2,2),(2,1,2),(2,2,1),(3,1,1)$ i.e. total of 6 cases

As we know,

Probability of occurrence of an event $=\frac{\text { Total no.of Desired outcomes }}{\text { Total no.of outcomes }}$

Therefore, the probability of outcome whose sum is 5

$=\frac{6}{216}$

$=\frac{1}{36}$

Conclusion: Probability of getting a total of 5 when three dice are thrown is $\frac{1}{36}$

(ii) Total no.of cases will be $6 \times 6 \times 6=216$ (because each die can have values from 1 to 6)

Desired outcomes are those whose sum up to 5 . Desired outcomes are $(1,1,3),(1,3$, $1),(1,2,2),(2,1,2),(2,2,1),(3,1,1)(1,1,1),(1,1,2),(1,2,1),(2,1,1)$, i.e. total of 10 cases

As we know,

Probability of occurrence of an event

$=\frac{\text { Total no. of Desired outcomes }}{\text { Total no.of outcomes }}$

Therefore, the probability of outcome whose sum is at most 5

$=\frac{10}{216}$

$=\frac{5}{108}$

Conclusion: Probability of getting total of at most 5 when three dice are thrown is $\frac{5}{108}$