Probability Practice Test 2

Question : 25 of 37

Marks: +1, -0

Two dices are thrown together. Find theprobability that the sum of the digits on thefaces is neither 10 nor 11.

$7/36$
$22/35$
$1/6$
$31/36$
$5/36$

Solution:

Total outcomes = 6 × 6 = 36
Pair of numbers that the gives sum of 10 = (4, 6), (6, 4), (5, 5)
Pair of numbers that gives the sum of 11 = (5, 6), (6, 5)
So, total ways of getting the sum of (10) or (11) is 5 ways.
So, the favorable cases of not getting a sum of (10) nor (11) are = 36 - 5 = 31
∴ Required probability

= 31/36

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