**Section 6.3**

(4)

S={GG, GB, BG, BB}

E = {GG}

P(E) = n(E)/n(S) = 1/4.

(10)

S = {GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB}

E = {GBB, BGB, BBG}

P(E) = n(E)/n(S) = 3/8.

(14)

Five cards are dealt. What is the probability that they are all hearts?

The sample space consists of all possible hands that could be dealt, so it has
size n(S) = C_{52, 5} =

52! /(5!(52 - 5)!) = 52! / (5! 47!) = (52)(51)(50)(49)(48) / 5! = (52)(51)(10)(49)(2)
= 2598960.

Let E be the event of all 5 cards being hearts. There are C_{13, 5}
= 13! / (5!(13-5)!) = 13! / (5! 8!) = 1287 ways of selecting 5 hearts from 13
hearts, so n(E) = 1287

Thus, P(E) = n(E)/n(S) = 1287 / 2598960 = 33 / 666400 ~ 0.000495168.

(18)

Call the candidates A, B, and C.

Any one could win, so the sample space, which consists of all possible outcomes, is S = {A, B, C}.

The probability assignment is P(A) = 2/5, P(B) = 2/5, and P(C) = 1/5.

(24)

Notice that n(S) = 36 in this problem. The ways the dice could add up to 8 are:

2 + 6, 6 + 2

3 + 5, 5 + 3

4 + 4

Call this event (i.e. the dice adding up to 8) E.

Then P(E) = n(E)/n(S) = 5/36.

(26)

Again, the sample space S has 36 elements. Here's how the sum could be greater
than 8:

3 + 6, 6 + 3

4 + 5, 5 + 4

4 + 6, 6 + 4

5 + 5

5 + 6, 6 + 5

6 + 6

Call this event (i.e. adding up to more than 8) E.

Then P(E) = n(E)/n(S) = 10/36 = 5/18.

(38)

Call the coins A, B and C, where C is the one with two heads. Denote an outcome
as, say, HTH, where the first letter is the outcome of A, the second is the
outcome of B, and the third is the outcome of C.

Then, S = {HHH, HTH, THH, TTH}.

The event E corresponding to two heads is E = {HTH, THH}.

Thus, P(E) = n(E)/n(S) = 2/4 = 1/2.

(40)

Set this up just like the previous one, except this time the event is E = ∅, because there can never be 0 heads because C will always come up heads. Thus P(E) = n(E)/n(S) = 0/4 = 0.

(42)

This time E = {TTH}, so P(E) = n(E)/n(S) = 1/4.

(58)

Five cards are dealt. What is the probability of dealing a 2, 3, 4, 5, and 6,
all of the same suit?

The sample space consists of all possible hands that could be dealt, so it has
size n(S) = C_{52, 5}.

The event E has just 4 things (i.e. 4 hands) in it, namely

E=**{** {2♥ 3♥ 4♥ 5♥ 6♥}, {2♦ 3♦
4♦ 5♦ 6♦}, {2♣ 3♣ 4♣ 5♣ 6♣},
{2♠ 3♠ 4♠ 5♠ 6♠}** }**

Therefore n(E) = 4, and

P(E) = n(E)/n(S) = 4/ C_{52, 5} = 4 / 2598960 = 1 / 649740 ~ 0.000001539.

(60)

Five cards are dealt. What is the probability of dealing 2 Kings and 3 Aces?

The sample space is the same as in the previous problem, so, as before, n(S)
= C_{52, 5}.

Think of forming the event E in two operations. First, select 2 of the 4 kings;
you can do this C_{4, 2} = 6 ways. Second, select 3 of the 4 aces. there
are C_{4, 3} = 4 ways to do this. Thus, by the multiplication principle,
n(E) = (6)(4) = 24.

Therefore P(E) = n(E)/n(S) = 24/ C_{52, 5} = 24 / 2598960 = 1 / 108290
~ 0.000009234.