Question

Let A, B and C be mutually independent events of a probability space (Ω, F, P), such that P(A) = P(B) = P(C) = 1 4 . Compute P((Ac ∩ Bc ) ∪ C). b) [4 points] Suppose that in a bicycle race, there are 19 professional cyclists, that are divided in a random manner into two groups. One group contains 10 people and the other group has 9 people. What is the probability that two particular people, let’s say Alex and Bob, will be in the same group ?

Answer 1

a.

Given that A, B and C are mutually independent events

Also P(A) = P(B) = P(C) = ¼ = 0.25

Please note the following formulae which will be used for solving the problem:

P(A∩B) = P(A) + P(B) - P(A ∪ B)

P(Ac) = 1- P(A)

So P(Ac) = P(Bc)= P(Cc) = 1-1/4 = ¾=0.75

If A, B and C are mutually independent events, the following properties holds:

1. P(A∩B) = P(A) * P(B)

2. P(A ∩B∩C) = P(A)* P(B) * P(C)

3. Ac, Bc and Cc are also mutually independent events

Which implies

P(Ac∩Bc) = P(Ac) * P(Bc)

P(Ac ∩Bc∩Cc) = P(Ac)* P(Bc) * P(Cc)

Now required is P((Ac ∩ Bc ) ∪ C).

= P(Ac ∩ Bc ) + P(C) - P((Ac ∩ Bc ) ∩ C)

= P(Ac)* P(Bc) + P(C) – [P(Ac)* P(Bc)* P(C)]

Substituting the values, we get

= 3/4*3/4 + ¼ - (3/4*3/4*1/4)

= 0.5625 + 0.25 – 0.140625

= 0.6718

b.

Required probability = Favorable outcomes/ overall outcomes

Calculations to find overall outcomes:

We have to divide the members into 2 groups with 10 and 9 members in each group. This is also equal to a selection of a single group of 10 members (since the remaining 9 members will automatically form another group)

So overall outcomes = Selecting 10 members randomly from a group of 19 members

                                     = 19C10

Calculations to find Favorable outcomes:

We have to divide into two groups (9 and 10 members)

There can be two possibilities namely Alex and Bob can be in a group of 10 or can be in a group of 9

So there are two cases

Case 1: Consider that Alex and Bob are in the group of 10.

Consider the number of ways of selecting 10 members from a group of 19 such that it contains Alex and Bob:

So there are a total of 19 members. Since 2 particular people, Alex and Bob should be in the same group. Let’s group them, considering them as one person say P.

Now the required group is an automatic selection of P and selecting 8 other members from the remaining group of 17. This is given by 17C8

Case 2: Consider that Alex and Bob are in the group of 9.

Consider the number of ways of selecting 9 members from a group of 19 such that it contains Alex and Bob:

So there are a total of 19 members. Since 2 particular people, Alex and Bob should be in the same group. Let’s group them, considering them as one person say P.

Now the required group is an automatic selection of P and selecting 7 other members from the remaining group of 17. This is given by 17C7

So overall favorable outcomes are given by 17C8 + 17C7

Hence required probability = Favorable outcomes/ overall outcomes

= (17C8 + 17C7)/19C10

=9/19

= 0.47