Jodie tosses a biased coin and throws two tetrahedral dice.The probability that the coin shows a head is. Each of the dice has four faces, numbered 1, 2, 3 and 4. Jodie’s score is calculated from the faces that the dice lands on, as follows:if the coin shows a head, the two numbers from the dice are added together;if the coin shows a tail, the two numbers from the dice are multiplied together.Find the probability that the coin shows a head given that Jodie ’s score is 8 .
＝1/5
(1/3)*(1/16)= 1/48
These events are actually not independent. Instead they're also not mutually exclusive, but as I've thought for a while it is more likely to be a combination of two possible outcomes with nothing in common. Because, there is a condition stated at the first place.
To find the probability that the coin shows a head given that Jodie's score is 8, we need to use conditional probability.
Let's break down the problem step by step:
Step 1: Determine the possible ways to get a score of 8 using two tetrahedral dice.
Jodie's score can be 8 in the following ways:
- 2+6
- 3+5
- 4+4
- 5+3
- 6+2
Step 2: Calculate the probabilities of getting each of these scores.
The probability of rolling each number on a tetrahedral die is 1/4, since there are four faces numbered 1, 2, 3, and 4, and each face has an equal chance of landing. Therefore, the probability of rolling each of the possible score combinations is as follows:
- P(2+6) = (1/4) * (1/4) = 1/16
- P(3+5) = (1/4) * (1/4) = 1/16
- P(4+4) = (1/4) * (1/4) = 1/16
- P(5+3) = (1/4) * (1/4) = 1/16
- P(6+2) = (1/4) * (1/4) = 1/16
Step 3: Calculate the probability that Jodie's score is 8.
To find the probability that Jodie's score is 8, we add up the probabilities of each of the possible score combinations:
P(score is 8) = P(2+6) + P(3+5) + P(4+4) + P(5+3) + P(6+2)
= (1/16) + (1/16) + (1/16) + (1/16) + (1/16)
= 5/16
Step 4: Use Bayes' theorem to calculate the probability that the coin shows a head given that Jodie's score is 8.
Bayes' theorem states that the conditional probability of event A given event B can be calculated as:
P(A|B) = (P(B|A) * P(A)) / P(B)
In this case, event A is the coin showing a head, and event B is Jodie's score being 8. So we can write:
P(H|score is 8) = (P(score is 8|H) * P(H)) / P(score is 8)
We already calculated P(score is 8) to be 5/16. Now we need to calculate P(score is 8|H), the probability of Jodie's score being 8 given that the coin shows a head. Since the coin shows a head, we need to sum the possible score combinations:
P(score is 8|H) = P(2+6|H) + P(3+5|H) + P(4+4|H) + P(5+3|H) + P(6+2|H)
When the coin shows a head, the numbers on the dice are added together. So:
P(2+6|H) = (1/4) * (1/4) = 1/16
P(3+5|H) = (1/4) * (1/4) = 1/16
P(4+4|H) = (1/4) * (1/4) = 1/16
P(5+3|H) = (1/4) * (1/4) = 1/16
P(6+2|H) = (1/4) * (1/4) = 1/16
Therefore, P(score is 8|H) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 5/16
Finally, we need to calculate P(H), the probability that the coin shows a head. The problem doesn't give this information, so we can assume it is biased. Let's say the probability of the coin showing a head is p.
Putting it all together, we have:
P(H|score is 8) = (P(score is 8|H) * P(H)) / P(score is 8)
= (5/16 * p) / (5/16)
= p
Therefore, the probability that the coin shows a head given that Jodie's score is 8 is p, where p is the probability of the coin showing a head.
P(biased coin) = 1/2 = P(A)
P(dice) = 1/16 = P(B)
P(A U B) = 1/2 + 1/16 = 9/16
(work it out using the venn diagram is also possible)
You don't state what the prob of heads or tails are of the biased coin.
Just to show the method of my calculations, I will assume the prob(heads) = 3/5 and prob(tails) = 2/5
Make yourself an addition and a multiplication table for 1,2,3,4
In the 16 possible sums, many are alike, there is only 1 sum of 8
to obtain that 8 , we have a prob of
(3/5)(1/16) = 3/80
in the multiplication table I see two 8's
to obtain one of those 8's, we have a prob of
(2/5)(2/16) = 1/20
So the prob of your event = 3/80 + 1/20 = 7/80
Change the calculations to reflect the actual probability of heads/tails of your biased coins