Six dice are thrown 729 times. How many times do you expect atleast three dice to show 5 or 6.....
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1 Answers
1Let the probability of occurrence of the event exactly in 3 dice be referred to as P (E, 3).
Let the event be, getting 4 or 5 when a die is thrown.
Probability of occurrence of the event in a single throw is p = 1/6 + 1/6= 1/3
q = 1 - p = 1 - 1/3= 2/3
Probability of occurrence of the event in at least 3 dice = P (E, 3) + P (E, 4) + P (E, 5) + P (E, 6)
This is a binomial situation and hence using the formula P (E, r) = nCr · pr · qn - r
The required probability = (6C3)(1/3)3 (2/3)3 + (6C4)(1/3)4(2/3)2 + (6C5)(1/3)5(2/3) + (6C6)(1/3)6
Required probability = (1/729) [(20 × 8) + (15 × 4) + (6 × 2) + 1] = 233/729 [Simplify.]
Out of 729 throws, the number of times atleast 3 dice will show a 4 or 5 is 729 × 233/729 = 233.
So, we can expect at least 3 dice to show a 4 or 5 for 233 times.