Bob has two coins, A and B, in front of him. The probability of Heads at each toss is p=0.5 for coin A and q=0.9 for coin B.
Bob chooses one of the two coins at random (both choices are equally likely).
He then continues with 5 tosses of the chosen coin; these tosses are conditionally independent given the choice of the coin.
Let:
Hi: the event that Bob's ith coin toss resulted in Heads;
N: the number of Heads in Bob's coin tosses.
1. For i∈{0,1,…,5}, pN(i), the pmf of N, is of the form
1/2(5 C a)b^5+c(5 C d) q^e(1−q)^f.
Find the coefficients a,b,…,f. Your answer can be either a number or an expression involving i.
a = ?
b = ?
c = ?
d = ?
e = ?
f = ?
To find the coefficients a, b, c, d, e, and f for the pmf of N, we need to consider the different cases for the number of heads in Bob's tosses.
For i∈{0,1,…,5}, the variable i represents the number of heads. Let's break down each case:
For i = 0:
In this case, all 5 tosses resulted in tails. The probability of getting tails in a single toss, given that Bob chose coin A, is (1 - p) = 0.5. Since Bob equally likely chooses coin A or B, the probability of choosing coin A is 0.5. So, the probability of getting 5 tails when choosing coin A is (0.5^5).
For i = 1:
In this case, exactly 1 toss resulted in heads. The probability of getting heads in a single toss, given that Bob chose coin A, is p = 0.5. Therefore, the probability of getting 1 head when choosing coin A is (5 C 1) * (0.5^1) * (0.5^4).
For i = 2:
In this case, exactly 2 tosses resulted in heads. The probability of getting heads in a single toss, given that Bob chose coin B, is q = 0.9. Therefore, the probability of getting 2 heads when choosing coin B is (5 C 2) * (0.9^2) * (0.1^3).
For i = 3:
In this case, exactly 3 tosses resulted in heads. The probability of getting heads in a single toss, given that Bob chose coin B, is q = 0.9. Therefore, the probability of getting 3 heads when choosing coin B is (5 C 3) * (0.9^3) * (0.1^2).
For i = 4:
In this case, exactly 4 tosses resulted in heads. The probability of getting heads in a single toss, given that Bob chose coin B, is q = 0.9. Therefore, the probability of getting 4 heads when choosing coin B is (5 C 4) * (0.9^4) * (0.1^1).
For i = 5:
In this case, all 5 tosses resulted in heads. The probability of getting heads in a single toss, given that Bob chose coin B, is q = 0.9. Therefore, the probability of getting 5 heads when choosing coin B is (5 C 5) * (0.9^5) * (0.1^0).
Now, let's summarize the coefficients for each case:
For i = 0:
a = 0 (no heads)
b = 5 (total number of tosses)
c = 0 (no heads)
d = 0 (no heads)
e = 4 (remaining tails after a heads)
f = 1 (remaining tails after a heads)
For i = 1:
a = 1 (one head)
b = 1 (single toss)
c = 4 (remaining tails after a head)
d = 1 (one head)
e = 4 (remaining tails after a head)
f = 0 (no tails)
For i = 2:
a = 2 (two heads)
b = 2 (two tosses)
c = 3 (remaining tails after two heads)
d = 2 (two heads)
e = 3 (remaining tails after two heads)
f = 0 (no tails)
For i = 3:
a = 3 (three heads)
b = 3 (three tosses)
c = 2 (remaining tails after three heads)
d = 2 (two heads)
e = 3 (remaining tails after three heads)
f = 0 (no tails)
For i = 4:
a = 4 (four heads)
b = 4 (four tosses)
c = 1 (remaining tails after four heads)
d = 2 (two heads)
e = 3 (remaining tails after four heads)
f = 0 (no tails)
For i = 5:
a = 5 (five heads)
b = 5 (five tosses)
c = 0 (no remaining tails after five heads)
d = 2 (two heads)
e = 3 (no remaining tails after five heads)
f = 0 (no tails)
Therefore, the coefficients for the pmf of N are:
a = 0, 1, 2, 3, 4, 5
b = 5, 1, 2, 3, 4, 5
c = 0, 4, 3, 2, 1, 0
d = 0, 1, 2, 2, 2, 2
e = 4, 4, 3, 3, 3, 3
f = 1, 0, 0, 0, 0, 0