Jodie tosses a biased coin and throws two fair 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
numbers on the faces that the dice land 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.
.

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

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can you hlep me do homework?

step-by-step explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°