Question

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 .

Answer 1

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

Answer 2

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)

Answer 3

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.

Answer 4

＝1/5

Answer 5

(1/3)*(1/16)= 1/48