Two dice are thrown 120 times. Find the average number of times in which the number on the first die exceeds the number on the second die.
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Answer:
50
Step-by-step explanation:
Given Two dice are thrown 120 times. Find the average number of times in which, the number on the first die exceeds the number on the second die
A pair of dice is thrown. So S will be {1,2,3,4,5,6 } two times
Now n(S) = 36
Let R be the random variable.
Now favourable cases are (2,1), (3,1) (3,2) (4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (5,4) (6,1) (6,2) (6,3) (6,4) (6,5)
So p is the probability of occurrence that the number on first dice exceeds the number on the second dice will be 15/36 = 5/12
Now E(R) = mean = n p
= 120 x 5/12
= 50
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