Probability question please help thanks

A game is played with a fair six-sided die. A player throws this die and if the result is 2,3,4 or 5, that result is the player's score. If the result is 1 or 6, the player throws the die a second time and the sum of the 2 numbers resulting from both throws is the player's score. Events A and B are as follows:
A: the player's score is 5,6,7,8 or 9
B: the player has two throws.

Show that P(A)=1/3
Find P(A U B) and P(A N D)
N = intersects

Ans: P(A N D)=1/6
P(AUB) = 1/2

Please help with the solution thanks. and explanation if possible

The best way to approach this problem is with simple enumeration of all cases:

Two rolls ... score ... probability

2, ... 2 ... 1/6
1,1 .... 2 .... 1/36

3, ... 3 ... 1/6
1,2, ... 3 .... 1/36

4, ... 4 ... 1/6
1,3 ... 4 .... 1/36

5, ... 5 ... 1/6
1,4 ... 5 .... 1/36

1,5 ... 6 ... 1/36

1,6 ... 7 ... 1/36
6,1 ... 7 ... 1/36

6,2 ... 8 .... 1/36

6,3 ... 9 .... 1/36

6,4 ... 10 .... 1/36

6,5 ... 11 .... 1/36

6,6 ... 12 .... 1/36

P(score = 5,6,7,8,9) =
(1/36 + 1/6) + 1/36 + (1/36 + 1/36) + 1/36 + 1/36
= 12/36 = 1/3

A Union B:

P(A U B) =
A = score of 5,6,7,8,9 ... the only single roll case is first roll = 5.
All other cases require two rolls, which is event B.
B = two rolls ... which occurs when you roll 1 or 6.
Hence, on 3 of the 6 possible first rolls you are in A U B,
3/6 = 1/2 so P(A U B) = 1/2

A intersect B:

P(A ^ B) = the rolls (1,4), (1,5), (1,6), (6,1), (6,2), and (6,3).
that's 6 of the rolls, each has probability 1/36,
6/36 = 1/6

i thought brevity was the soul of wit !

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short working
------------------
................................. <----------------------- P(A ∩ B) --------------------------------->
P(A) = P(5 on #1) + [ P(1 on #1)*P(4-6 on #2) + P(6 on #1)*P(1-3 on #2) ]

= 1/6 + [1/6*1/2 + 1/6*1/2 ] = 1/3 <---------

P(A ∩ B) from above = 1/6 <--------

P(A U B) = P(A) + P(B) - P(A ∩ B) = 1/3 + 1/3 - 1/6 = 1/2 <--------