Question

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(MA-262 review) A fair six-sided die is rolled four times, and each result is recorded, in order. Determine (a) the probability that there are exactly two results (among the four) that are each a 3, and (b) the probability that the sum of the four results is 23. [Answers: 0.11574, 0.0030864.]

Answer 1

two three comes in (3311),(3312),(3314),(3315),(3316)(3322)(3324)(3325)(3326)(3344)(3345)(3346)(3355)(3356)(3366)

total permutation with 3311 is 4!/(2!2!) =6

total permutation with 3312 is 4!/2 =12

total permutation with 3314 is 4!/2 =12

total permutation with 3315 is 4!/2 =12

total permutation with 3316 is 4!/2 =12

total permutation with (3324)(3325)(3326)3345)(3346)(3356) = 12*6 =72

total permutation with (3322)(3344)(3355)(3366) = 6*4 = 24

so, total favorable cases=6+12+12+12+12+72+24 = 150

total outcomes when four die are rolled = 6^4 = 1296

P(exactly two three) = 150/1296=0.11574

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sum of 23 is made of (5+6+6+6)

total permutation with(5,6,6,6) = 4!/3! = 4

so, P(sum of 23) =4/1296 = 0.0030864

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