Probability Questions

I'm having a bit of a problem solving these few questions on probability, I hope some of u can help me out.

1. Two cubical fair dice are thrown, one red and one blue. The scores on their faces are added together. Determine which, if either, is greater:

a) The probability that the total score will be 10 or more given that the red dice shows a 6.
b) The probability that the total score will be 10 or more given that at least one of the dice shows a 6.

2. The probability that an event A occurs is P (A) = 0.3. The event B is independent of A and P (B) = 0.4.

a) Calculate P(A or B or both occur)

Event C is defined to be the event that neither A nor B occurs.

b) Calculate P (C/A'), where A' is the event that A does not occur.

Reply 1

danniella

4

Hello!

Question 2:
a) P(A) = 0.3, P(B) = 0.4
As A and B are independent,
P(A and B) = P(AnB) = 0.3x0.4 = 0.12

Probability that either A or B or both is called A union B (AuB)
So P(AuB) = P(A) + P(B) - P(AnB)
(Either A on its own, or B on its own, but then I've counted "A and B together" twice (once in A and once in B) so subtract one copy of "A and B together")

P(AuB) = P(A) + P(B) - P(AnB)
P(AuB) = 0.4 + 0.3 - 0.12
P(AuB) = 0.58

b) P(C given that A doesn't happen) is P(C/A')
P(C/A') = P(C n A') / P(A')

(i.e. the probability that C happens given that A doesn't happen is the probability of both C happening and A doesn't, divided by the probability that A doesn't happen)

P(C happens and A doesn't) = 0.42
( 1 - your answer to B, i.e. 1-the probability that either A or B or both happens)

P(A doesn't happen) = 1-0.3 = 0.7

So P(C/A') = P(C n A') / P(A')
and so you get
P(C/A') = 0.42/0.7 = 6/10 = 0.6

love danniella

Reply 2

Civ-217

4

andrewlee89

1. Two cubical fair dice are thrown, one red and one blue. The scores on their faces are added together. Determine which, if either, is greater:

a) The probability that the total score will be 10 or more given that the red dice shows a 6.
b) The probability that the total score will be 10 or more given that at least one of the dice shows a 6.

Blue
6| 7 | 8 | 9 | 10| 11| 12 |
5| 6 | 7 | 8 | 9 | 10| 11 |
4| 5 | 6 | 7 | 8 | 9 | 10 |
3| 4 | 5 | 6 | 7 | 8 | 9 |
2| 3 | 4 | 5 | 6 | 7 | 8 |
1| 2 | 3 | 4 | 5 | 6 | 7 |
--------------------------
1 | 2 | 3 | 4 | 5 | 6 | Red

These are the outcomes.

Now
P(red dice shows a red) = 1/6
P(at least one of them shows a 6) = 11/36
P(total score >= 10) = 1/6

a)
P(total score>=10 | Red dice shows a 6)
= P(total score >=10 and red dice shows a 6) / P(red dice shows a 6)
= 3/36 / (1/6)
= 1/2

b)P(total score>=10 | at least one of them shows a 6)
= P(total score >=10 and at least one of them shows a 6) / P(at least one of them shows a 6)
= 5/36 / (1/6)
= 5/6

Reply 3

kai68

Original post by Civ-217

Blue
6| 7 | 8 | 9 | 10| 11| 12 |
5| 6 | 7 | 8 | 9 | 10| 11 |
4| 5 | 6 | 7 | 8 | 9 | 10 |
3| 4 | 5 | 6 | 7 | 8 | 9 |
2| 3 | 4 | 5 | 6 | 7 | 8 |
1| 2 | 3 | 4 | 5 | 6 | 7 |
--------------------------
1 | 2 | 3 | 4 | 5 | 6 | Red

These are the outcomes.

Now
P(red dice shows a red) = 1/6
P(at least one of them shows a 6) = 11/36
P(total score >= 10) = 1/6

a)
P(total score>=10 | Red dice shows a 6)
= P(total score >=10 and red dice shows a 6) / P(red dice shows a 6)
= 3/36 / (1/6)
= 1/2

b)P(total score>=10 | at least one of them shows a 6)
= P(total score >=10 and at least one of them shows a 6) / P(at least one of them shows a 6)
= 5/36 / (1/6)
= 5/6

Good try their pal but their's a slight error,

P(total score >=10 and at least one of them shows a 6) / P(at least one of them shows a 6)
=5/36/11/36
=5/11 or 0.454......

This is because the P(atleast one of the dies shows a six)= 11/36
i.e (R1,b6),(R2, b6)........all the way to (R6, b6)
and (R5, b6), (R4 b6), all the way to (R1, b6)
Happy to help :wink: