Question

Problem 18: Roll a die four times. What is the probability of

(a) Getting a pair?

(b) Getting three of a kind?

(c) Getting four of a kind?

(d) Getting two pairs?

Four distinct faces?

Answer 1

a)

total number of ways without any constraints = 6^4

number of ways to get a paire = total number - number of ways that all are different

= 6^4 - 6 * 5 * 4*3 = 6^4 - 360

hence

required probabiliy = (6^4 - 360)/6^4

= 1 - 360/6^4

=13/18

b)

getting three of a kind

= 6C1 * 5C1 * 4 = 120   { 6 ways to select which will be repeated 3 times, 5 ways to select other and there are 4 ways to arrage AAAB type}

hence required probability = 120/6^4

c) getting 4 of a kind

= 6/6^4 = 1/6^3

d) getting two pairs

= 6C2 * 4!/(2!   *2!)

= 15 * 6 =90

{6C2 to select which two pairs , 4!/(2! *2!) to arrange AABB type}

hence 90/6^4

e) 4 distinct face

= 6*5*4*3 /6^4 { similar to A) part }