John has three coins in his pocket. One is fair, one has two heads, and the third is unbalanced, showing head 60% of the time. John grabs a coin at random and tosses it. Find the probability that:
1) The coin shows a head
2) The coin is regular, give that the toss produced a head
3) The coin is regular, given that the toss produced a tail
1. Probability of showing a head:
Coins are chosen at random, so each coin has a probability of (1/3) of being selected.
probability for a head:
fair coin: (1/3)(1/2)=1/6
two heads: (1/3)(1)=1/3
unbalanced: (1/3)(0.6)=1/5
Add up the three probabilities to get 7/10.
Conditional probabilty:
For events A and B, P(A|B) stands for probability of A happening given that B is true. In numerical terms,
P(A|B)=P(A∩B)/P(B)
2. Let
A=coin is regular
B=produced a head
P(A|B)=P(A∩B)/P(B)
P(A∩B)=[(1/3)(1/2)]=1/6 as calculated above
P(B)=7/10 as calculated above
P(A|B)=(1/6)/(7/10)
=5/21
3. similar to (2), left as exercise for you.
To answer these questions, we can apply conditional probability. Let's denote the events:
A: The coin shows a head
B: The coin is regular
1) The probability that the coin shows a head (P(A)) can be calculated using the law of total probability. We will consider the individual probabilities based on the type of coin John selected.
P(A) = P(A|B1) * P(B1) + P(A|B2) * P(B2) + P(A|B3) * P(B3)
P(A|B1) = 1/2 (as the fair coin has heads 50% of the time)
P(B1) = 1/3 (as there are three coins)
P(A|B2) = 1 (as the coin with two heads will always show a head)
P(B2) = 1/3
P(A|B3) = 0.6 (as the unbalanced coin shows heads 60% of the time)
P(B3) = 1/3
Plugging in these values, we get:
P(A) = (1/2) * (1/3) + 1 * (1/3) + 0.6 * (1/3)
P(A) = 0.5/6 + 3/6 + 0.6/6
P(A) = 4.1/6
P(A) ≈ 0.6833
Therefore, the probability that the coin shows a head is approximately 0.6833.
2) To calculate the probability that the coin is regular, given that the toss produced a head (P(B|A)), we can use Bayes' theorem:
P(B|A) = (P(A|B) * P(B)) / P(A)
P(A|B) = 1/2 (as the fair coin has heads 50% of the time)
P(B) = 1/3
P(A) = 4.1/6 (as calculated in question 1)
Plugging in these values, we get:
P(B|A) = (1/2) * (1/3) / (4.1/6)
P(B|A) ≈ 0.0976
Therefore, the probability that the coin is regular, given that the toss produced a head, is approximately 0.0976.
3) Now, let's calculate the probability that the coin is regular, given that the toss produced a tail (P(B|not A)). Here, "not A" represents the event that a tail is observed.
P(B|not A) can be calculated using Bayes' theorem as well:
P(B|not A) = (P(not A|B) * P(B)) / P(not A)
P(not A|B) = 1/2 (as the fair coin has tails 50% of the time)
P(B) = 1/3
P(not A) = 1 - P(A) ≈ 1 - 0.6833 ≈ 0.3167 (as calculated in question 1)
Plugging in these values, we get:
P(B|not A) = (1/2) * (1/3) / 0.3167
P(B|not A) ≈ 0.1664
Therefore, the probability that the coin is regular, given that the toss produced a tail, is approximately 0.1664.