Question

You have a large barrel full of coins. 5 percent of the coins in the barrel are “type X” and 95 percent are “type Y.” When you flip them, type X coins come up heads 90 percent of the time and tails 10 percent of the time. Type Y coins come up heads 3 percent of the time and tails 97 percent of the time.

Suppose you do a three part experiment: (i) Take a coin at random from the barrel [in this context, choosing a coin “at random” from the barrel means that there is a 5 percent chance you will select a type X coin and a 95 percent chance you will select a type Y coin] , (ii) flip the coin you selected once and observe whether it comes up heads or tails, and (iii) flip the same coin a second time and observe whether it comes up heads or tails.

a) Represent this experiment in a tree, showing all the possible outcomes, along with the probability of each outcome. [For example, one possible outcome is "the coin is type Y, the first flip is tails, and the second flip is heads," which you could abbreviate as YTH.]

b) What is the probability that the two tosses of the coin do not give the same result? (That is, what is the probability that either the first toss is heads and the second toss is tails, or the first toss is tails and the second toss is heads?)

c) If the two tosses of the coin do not give the same result, what is the probability that the coin chosen at random is type Y?

d) Define event A to be the set of all outcomes of this experiment in which the first flip comes up Heads. Define event B to be the set of all outcomes of this experiment in which the second flip comes up Heads. Are A and B independent events? Justify your answer carefully, using an argument based on the definition of statistical independence, not just intuition.

Answer 1

We start by listing all the outcomes and their respective probabilities. Each result's probability is mutiplied with the other stages since there is independent selection of type and the outcomes of the toss

Type	1 toss	2 toss	Outcome	P(type)	P(1 toss)	P(2 toss)	P(outcome)
X	H	H	XHH	0.05	0.9	0.03	0.00135
X	H	T	XHT	0.05	0.9	0.97	0.04365
X	T	H	XTH	0.05	0.1	0.03	0.00015
X	T	T	XTT	0.05	0.1	0.97	0.00485
Y	H	H	YHH	0.95	0.9	0.03	0.02565
Y	H	T	YHT	0.95	0.9	0.97	0.82935
Y	T	H	YTH	0.95	0.1	0.03	0.00285
Y	T	T	YTT	0.95	0.1	0.97	0.09215

Note : The sum of all probabilities is '1' which fulfils the condition of being a probabiltiy distribution.

a) Represent this experiment in a tree, showing all the possible outcomes, along with the probability of each outcome. [For example, one possible outcome is "the coin is type Y, the first flip is tails, and the second flip is heads," which you could abbreviate as YTH.]

b) What is the probability that the two tosses of the coin do not give the same result? (That is, what is the probability that either the first toss is heads and the second toss is tails, or the first toss is tails and the second toss is heads?)

That mean ignore the outcomes of TT and HH irrespective of 'x' and 'Y'

which then are

Outcome	P(outcome)
XHT	0.04365
XTH	0.00015
YHT	0.82935
YTH	0.00285
Total	0.876

Probability that the two tosses of the coin do not give the same result is 0.876

c) If the two tosses of the coin do not give the same result, what is the probability that the coin chosen at random is type Y?

This is an example of conditional probability

P( Type Y | diferent result) =

=

From the above tables we can get the probabilites and we put in our formula

= 0.95

Probability that the coin chosen at random is type Y is 0.95.

d) Define event A to be the set of all outcomes of this experiment in which the first flip comes up Heads. Define event B to be the set of all outcomes of this experiment in which the second flip comes up Heads. Are A and B independent events? Justify your answer carefully, using an argument based on the definition of statistical independence, not just intuition.

The events are independent when

P(A B) = P(A) * P(B)

Where 'A' is the event that heads is on the first toss

and 'B' is the event that heads is on the second toss

P(A B) will be the event that heads is on both the tosses.

The outcomes therefore are = P( XHH) + P(YHH)

= 0.027

P(A) : We take all the outcomes with H in the 1st toss ignoring the type and 2nd outcome

Outcome	P(outcome)
XHH	0.00135
XHT	0.04365
YHH	0.02565
YHT	0.82935
	0.9

P(A) = 0.9

P(B) : We take all the outcomes with H in the 2nd toss ignoring the type and 1st outcome

Outcome	P(outcome)
XHH	0.00135
XTH	0.00015
YHH	0.02565
YTH	0.00285
	0.03

P(B) = 0.03

P(A) * P(B) = 0.027

Since

We can say that events 'A' and 'B' are independent.