Three dice are rolled. What is the probability that the first gives a one and th

Question

Three dice are rolled. What is the probability that the first gives a one and the other two give sixes? Explain in terms of probability rules.

Consider a standard deck of 52 cards. If three cards are chosen, what is the probability that the first is a jack while the second and third are kings?

Five dodecahedral (i.e. 12-sided) dice are rolled.
(a) What is the probability, p, that when one is rolled, the number is a multiple of three?
(b) What is the probability of getting exactly two multiples of three?
(c) What is the probability of getting > 2 multiples of three?
(d) If 500 such dice are rolled, what is the probability of getting 100 multiples of three?

Explanation / Answer

1. three dice are rolled
Total possible outcomes are 216
Required outcome is possible in 3 ways

Hence required probabiltiy = 3/216 = 0.0139

2. 3 cards are chosen
3 cards from a deck of 52 cards can be drawn in 52
C3 = 22100 ways

1st is jack is possible in 4 ways
2nd is king is possible in 4 ways and 3rd is king is possible in 3 ways

Hence required probability = *4*4*3)/52C3 = 0.0022

3. Five dodecahedral dice
a) p = 4/12 = 0.3333

b)
P(Exactly two multiples of 3) = 5C2 * (0.3333)^2 * (1-0.3333)^3 = 0.3292

c)
P(> 2 multiples of 3) = 1 - P(X = 0) - P(X=1) - P(X=2)
= 1 - 5C0 * (0.3333)^0 * (1-0.3333)^5 - 5C1 * (0.3333)^1 * (1-0.3333)^4 - 5C2 * (0.3333)^2 * (1-0.3333)^3 = 0.2098

d)
binomial apporximation to normal
mean = 0.3333*500 = 166.65
std. dev. = sqrt(0.3333*(1-0.3333)*500) = 10.5407

P(X <= 100)
= P(z <= (100 - 166.65)/10.5407)
= P(z <= -6.3231)
= 0

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