Question

Please give detailed solution for this

Coin 1 comes up heads with probability .3, whereas coin 2 comes up heads with probability .6. A coin is randomly chosen and flipped 10 times.

(a) Find the probability the first flip lands heads.

(b) Find the expected number of heads in the 10 flips.

(c) Find the probability that there are a total of 7 heads.

Answer 1

Since the coin is randomly chosen, the probability of being chosen for each of them is 1/2. Let A be the event of choosing coin 1 and B be the event of choosing coin 2, so that P(A) = 1/2 and P(B) = 1/2. Now, any event C can be viewed as the union of two disjoint sets, i.e. . Using the Bayes' theorem,

a.

Here C= first flip lands Head.

P(C|A) = 0.3, P(C|B) = 0.6

So, P(C) = 0.3*1/2 +0.6*1/2 = 0.45

b. C = number of heads in 10 flips

E(C) = (definition of expectation of a random variable)

Given a coin, the probability of the number of heads in 10 flips is given by a binomial probability distribution.

The above expression contains two terms in addition, each of which is the expectation of Bin(10,0.3) and Bin(10,0.6) respectively. It is a known result that the expectation of a Bin(n,c) = nc. Using this result,

E(C) = 0.5 * (10*0.3+10*0.6) = 0.5 * (3+6) = 4.5

c.

C = 7 out of 10 flips give heads

Hence, P(C) =