A bag contains 5 coins, of which 4 are fair; the remaining coin has a head on both sides. If a coin is selected at random from the bag and tossed three times, construct a probability tree to determine the probability that heads will occur at least twice.
A)Pr(at least two heads) =
B)Pr(at least two heads) =
C)Pr(at least two heads) =
D)Pr(at least two heads) =
E)Pr(at least two heads) =

Question

A bag contains 5 coins, of which 4 are fair; the remaining coin has a head on both sides. If a coin is selected at random from the bag and tossed three times, construct a probability tree to determine the probability that heads will occur at least twice. ​
A)Pr(at least two heads) = 
B)Pr(at least two heads) = 
C)Pr(at least two heads) = 
D)Pr(at least two heads) = 
E)Pr(at least two heads) =

Quizplus · Accepted Answer

The probability tree can be constructed as follows:
First Branch:
P(Fair) = 4/5, P(Biased) = 1/5
Second Branch:
P(Head) = 1/2 for Fair, P(Head) = 1 for Biased
P(Tail) = 1/2 for Fair, P(Tail) = 0 for Biased
Third Branch:
P(Head) = 1/2 for Fair, P(Head) = 1 for Biased
P(Tail) = 1/2 for Fair, P(Tail) = 0 for Biased
The probability of getting at least two heads can be calculated by adding the probabilities of getting 2 heads and 3 heads.
For Fair coin:
P(2 heads) = (4/5)(1/2)(1/2)(1/2) = 1/16
P(3 heads) = (4/5)(1/2)(1/2)(1/2) = 1/16
For Biased coin:
P(2 heads) = (1/5)(1)(1)(1/2) = 1/10
P(3 heads) = (1/5)(1)(1)(1) = 1/5
Using the Law of Total Probability, the total probability of getting at least two heads can be calculated as:
P(at least two heads) = P(Fair)P(at least two heads|Fair) + P(Biased)P(at least two heads|Biased)
= (4/5)(1/16+1/16) + (1/5)(1/10+1/5) = 11/40
Therefore, the correct answer is C.

A Bag Contains 5 Coins, of Which 4 Are Fair;