Question

A fair coin is tossed 3 times in a row. Use Kolmogorov theory and identify the sample space S using the

2
3

arrangements for 3 tosses. i) Clearly identify all the sets of
S
and assign a probability assuming equiprobable events. ii) Then compute the probability associated to the following events (sets) (do not use combinatorics)
A
0
​
={0
heads after 3 tosses
}
A
1
​
={1
head after 3 tosses
}
A
2
​
={2
headsafter 3 tosses
}
A
3
​
={3
heads after 3 tosses
}
. iii) Then, show that
S=A
0
​
∪A
1
​
∪A
2
​
∪A
3
​

and that the partition events
A
j
​

are mutually exclusive, i.e. they are disjoint sets. iv) Then compute the probabilities
P(A
j
​
)
using just combinatorics.

Answer 1

i) The sample space S for the experiment of tossing a fair coin 3 times in a row using arrangements can be written as:

S = {(H,H,H), (H,H,T), (H,T,H), (T,H,H), (T,T,H), (T,H,T), (H,T,T), (T,T,T)}

where H denotes heads and T denotes tails.

All the sets of S and their corresponding probabilities assuming equiprobable events are:

A0 = {TTT} with probability P(A0) = 1/8

A1 = {(H,T,T), (T,H,T), (T,T,H)} with probability P(A1) = 3/8

A2 = {(H,H,T), (H,T,H), (T,H,H)} with probability P(A2) = 3/8

A3 = {HHH} with probability P(A3) = 1/8

ii) The probability associated with the following events (sets) are:

P(A0) = 1/8

P(A1) = 3/8

P(A2) = 3/8

P(A3) = 1/8

iii) To show that S = A0 ∪ A1 ∪ A2 ∪ A3, we need to show that every element in S belongs to at least one of the sets A0, A1, A2, A3, and no element in S belongs to more than one of these sets.

Since S contains all possible outcomes of the experiment, it is clear that every element in S belongs to at least one of the sets A0, A1, A2, A3.

Moreover, we can see that each element in S contains exactly one of the following number of heads: 0, 1, 2, or 3. Therefore, no element in S can belong to more than one of the sets A0, A1, A2, A3. Hence, we have shown that S = A0 ∪ A1 ∪ A2 ∪ A3.

To show that the partition events A0, A1, A2, A3 are mutually exclusive, we need to show that they do not have any common elements, i.e., A0 ∩ A1 = A0 ∩ A2 = A0 ∩ A3 = A1 ∩ A2 = A1 ∩ A3 = A2 ∩ A3 = ∅.

It is clear that A0, A1, A2, A3 are mutually exclusive from their definitions.

iv) To compute the probabilities P(Aj) using combinatorics, we can use the formula:

P(Aj) = number of outcomes in Aj / total number of outcomes in S

The number of outcomes in Aj can be calculated using combinatorics. For example, to find the number of outcomes in A1, we can use the formula for the number of ways to choose k items from n items without replacement, which is given by the binomial coefficient:

n choose k = n! / (k!(n-k)!)

So, the number of outcomes in A1 can be calculated as follows:

Number of outcomes in A1 = 3 choose 1 = 3! / (1!(3-1)!) = 3

Using the same approach, we can find the number of outcomes in the other sets as:

Number of outcomes in A0 = 1 choose 0 = 1

Number of outcomes in A2 = 3 choose 2 = 3! / (2!(3-2)!) = 3

Number of outcomes inA3 = 1 choose 3 = 1

Using these values, we can calculate the probabilities P(Aj) as:

P(A0) = 1/8

P(A1) = 3/8

P(A2) = 3/8

P(A3) = 1/8

Note that these probabilities add up to 1, as expected.