A jar contains 3 coins. 2 are heads and tails. 1 is heads and heads.
You pick out at coin and toss it 3 times. You get three heads.
Question ; What is the probability of getting a head on the fourth toss ?
Solution:
To determine the probability of getting a head on the fourth toss after tossing three heads, we need to consider the coins in the jar: two coins are heads and tails (A and B), and one coin is heads and heads (C). The probability of selecting each coin is 1/3.
Given that we have already tossed three heads, we can use conditional probability. The only way to get three heads is if we have selected coin C (the heads and heads coin) or one of the other coins (A or B) and happened to get heads all three times. The probability of getting three heads with coin A or B is (1/2)^3 = 1/8 for each, and since there are two such coins, the total probability for A and B is 2 * (1/3) * (1/8) = 1/12.
The total probability of getting three heads is the sum of the probabilities for all coins: P(3 heads) = P(3 heads | C) + P(3 heads | A or B) = 1/3 + 1/12 = 5/12.
Now, we want to find the probability of having coin C given that we got three heads: P(C | 3 heads) = P(3 heads | C) * P(C) / P(3 heads) = (1/3) * (1) / (5/12) = 4/5.
Thus, the probability of getting a head on the fourth toss is: P(head on 4th toss | 3 heads) = P(head | C) * P(C | 3 heads) + P(head | A or B) * P(A or B | 3 heads) = 1 * (4/5) + (1/2) * (1/5) = 4/5 + 1/10 = 9/10.
Therefore, the probability of getting a head on the fourth toss is 9/10.
Comments (9)
I read a puzzle and its answer and cannot figure it out.
A jar contains 3 coins. 2 are heads and tails. 1 is heads and heads.
You pick out at coin and toss it 3 times. You get three heads.
Question ; What is the probability of getting a head on the fourth toss ?
The answer given was 90% ... I just don't get it
3/8 using a probability tree.
Could not be bothered to try the binomial theorem as I have no tables.
Me thinks it be 2/3.
Probability of getting 4 heads from the start:
P(A)=2*(1/3)*(1/16)+1/3=9/24
Probability of getting 3 heads from the start:
P(B)=2*(1/3)*(1/8)+1/3=10/24
Probability of getting 4 heads given that 3 heads flipped already:
P(A given B)=P(A)/P(B)=9/10
The probability of another head is 9/10
No need for probability trees, or even looking at the history of the first 3 throws!
The question only asks about the fourth toss.
There is
1/3 probability of coin A (with p(heads) = 1/2) being used
OR
1/3 probability of coin B (with p(heads) = 1/2) being used
OR
1/3 probability of coin C (with p(heads) = 1) being used
P(A OR B OR C) = P(A) + P(B) + P(C)
= (1/3 * 1/2) + (1/3 * 1/2) + (1/3 * 1)
= 2/3
bathfilms, yes you do need to look at the first three throws, because the first 3 heads tell you that it is more likely you are holding a head and head coin rather then a head and tail coin. In other words, the probabilities of having the three coins are not 1/3 each.
For example, if you started with 1 head and head coin, but 2 tail and tail coins, and you throw three heads, the probability of a 4th head is clearly 1, but by your calculation it would be 1/3.
This is a question in conditional probability, the solution is as I posted above.
Dear Dzallen,
Yes, thank you, I see your point of view, which is of course correct.
So, if the 1st three tosses were heads, but you had your eyes closed while so-doing, that would change the probability. It's a bit like Monty Hall's dilemma.
Surely the answer is 2/3. Label the coins A and B (each has a head & tail) and C for the two headed coin.
The probability of selecting coin A is 1/3. The probability of throwing a head regardless of previous throws is 1/2. The probabilit of a Head if Coin A is 1/3 multiplied by 1/2 = 1/6. The same applies for Coin B.
For coin C the probability is 1/3 to select it and then 1 to throw a head = 1/3. So the probability of a head is 1/6 or 1/6 or 1/3. For 'or' results you add the probabilities ie 1/6+1/6+1/3 =2/3
2/3
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