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Three dice are thrown together. Find the probability of getting a total of at least 6. |
$\frac{5}{108}$ $\frac{5}{18}$ $\frac{101}{108}$ $\frac{103}{108}$ |
$\frac{103}{108}$ |
The correct answer is Option (4) → $\frac{103}{108}$ If three dice are thrown, total sample space =216
The event of getting the sum less than 6 =1,1,1,1,1,2,1,2,1,2,1,1,1,2,2,2,2,1,2,1,2,1,1,3,3,1,1,1,3,1
Since the number of outcomes, in which the sum less than 6 is 10.
The probability of getting a total less than 6 =10 / 216
Therefore, the probability of getting a total of atleast 6 =1-10/216
The probability of getting a total of atleast 6 =103/108
Hence option D is the correct option. |
