2. Suppose we have two sets of dice. The first set contains four fair six-sided

Question

2. Suppose we have two sets of dice. The first set contains four fair six-sided (i.e. with numbers 1, 2,..., 6) dice and these dice are colored red. The second set also contains four six-sided dice (i.e. with numbers 1, 2, ..., 6) and these dice are colored blue. However, unlike the red dice (which are fair), each of the blue die are biased in the sense that numbers divisible by 3 (i.e. 3 and 6) are three times more likely than the other four numbers (which have the same chance of occurring). The experiment will consist of rolling either the red or blue dice independently of one another. (a) If the red dice are rolled, what is the probability of getting "four of a kind" (i.e. all dice have the same number)? (b) If the red dice are rolled, what is the probability of getting "two pairs" (e.g. two 2's and two 3's or two 4's and two 3's, but not four 2's)? (e) Solve part (a) assuming the blue dice are rolled. (d) Solve part (b) assuming the blue dice are rolled

Explanation / Answer

a.)

Total Possible Outcomes = 64 = 1296

number of ways dice have same faces = 6

Probability of getting all dice with same face = 6 / 1296 = 1 / 216

b.)

Possible pairs are = 6C2 = 15 pairs

But these pairs can also be arranged in 4! / 2! * 2! = 6 ways

Total Possible ways = 15 * 6 = 90

Hence Total Probability = 90 / 1296 = 15 / 216 = 5 / 72

Blue dice is biased so such 3 and 6 has 3 times probability that of other 4 numbers. So,

p + p + 3p + p + p + 3p = 1

p = 1 / 10

P(3 or 6) = 3 / 10

P(other 4 numbers) = 1 / 10

c.)

We need all 4 numbers same:

Probability of getting 1,2,4 and 5 on all 4 dices, P(1or2or3or4) = 4 * (1/10)4 = 4/10000

Probability of getting 3 or 6 on all 4 dices, P(3 or 6) = 2 * (3/10)4 = 162/10000

Total Probability = 166/10000 = 0.0166

d.)

Similarly as part b, 15 pairs are possible here as well

But there are 5 pairs which don't contain 3 or 6

Also each pair can be arranged in 6 ways

So, P(5 pairs without 3 or 6) = 5 * 6* (1/10)4 = 30 / 10000

There are 9 pairs which contain either 2 6's or 2 3's

P(9 pairs with 2 3's or 6's) = 9 * 6* (1/10)4 = 54 / 10000

1 pair has 2 3's and 2 6's(3,3,6,6)

P(1 pair with 2 3's and 2 6's) = 1 * 6 * 1/10000 = 6/10000

Total Probability = 90/10000 = 0.009

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