Amy plays a game with an ordinary, fair dice.
If she rolls: 1 she wins, 2 or 3 she loses, 4, 5 or 6 she rolls again.
When she has to roll again: if she rolls an even number she wins
if she rolls an odd number she loses.
a) Work out the probabilities that should be
First roll
Second roll
placed at P, Q, R and S on the tree diagram.
Give your answers as fractions in the form a/b
P
Q
2 or 3
P:
Q:
R:
S:
R
even
4, 5 or 6
(1)
(1)
b) Work out the probability that Amy will win
and the probability she will lose.
Give your answers as fractions in the form a/b
S
odd
P(win) =
P(lose) =
(2)
(2)

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adinaafsar adinaafsar · Answer 1 · 2021-01-28T10:53:28+01:00

Answer:

Ok so I’m stuck with this question too But I got the top part.

p=1/6

q=2/6

r=3/6

s=3/6

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AFOKE88 AFOKE88 · Answer 2 · 2021-10-19T19:37:51+02:00

A) The probabilities placed at P, Q, R and S on the tree diagram are;

P = 1/6; Q = 1/3; R = 1/4; S = 1/4

B) The probabilities of her winning or losing is;

Probability that she will win = 5/12

Probability that she will lose = 7/12

At point P; Only 1 is rolled.

Since there are 6 possible numbers, then Probability at P = 1/6

At point Q; either 2 or 3 can be rolled.

Since there are 6 possible numbers, then Probability at Q = 2/6 = 1/3

At point R; 4,5, or 6 can be rolled on first roll.

Thus, Probability on first roll = 3/6

At R on second roll if she gets even, probability on second roll = 3/6

At S on second roll if she gets odd, probability on second roll = 3/6

Thus;

Overall probability at R = 3/6 × 3/6 = 1/4

Overall probability at S = 3/6 × 3/6 = 1/4

Thus, probability that she will win = 1/6 + 1/4 = 5/12

Probability that she will lose are the probabilities at Q and S.

Probability that she will lose = 1/3 + 1/4 = 7/12

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