# probability - plz help

• Apr 26th 2010, 01:22 PM
dee2020
probability - plz help
1. If P(T) = P(Q) =3(T intersection Q) and P(T U Q) = 0.75
Find
(a) P(T intersection Q)
(b) P(T)
(c) P(Q')

2. Two dice are thrown. One die lands on 2. Find the probability that product of dice is
(a) exactly 6
(b) more than 5

Any help or hint much appreciated

Dee
• Apr 26th 2010, 01:29 PM
Plato
Quote:

Originally Posted by dee2020
1. If P(T) = P(Q) =3(T intersection Q) and P(T U Q) = 0.75
Find
(a) P(T intersection Q)
(b) P(T)
(c) P(Q')

2. Two dice are thrown. One die lands on 2. Find the probability that product of dice is
(a) exactly 6
(b) more than 5

1) $\displaystyle P(T\cup Q)=P(T)+P(Q)-P(T\cap Q)$

2) There are thirty-six possible pairs.
The event where the product equals six is $\displaystyle \{(2,3),(3,2)\}$
SO?
• Apr 26th 2010, 01:53 PM
dee2020
for 2nd question, do i have to use conditional probability formula
P(B|A) = P(B intersection A)/ P(A)
• Apr 26th 2010, 03:04 PM
Plato
Quote:

Originally Posted by dee2020
for 2nd question, do i have to use conditional probability formula
P(B|A) = P(B intersection A)/ P(A)

Eleven pairs contain a two. So $\displaystyle \frac{2}{11}$
• Apr 27th 2010, 04:30 AM
dee2020
Quote:

Originally Posted by Plato
Eleven pairs contain a two. So $\displaystyle \frac{2}{11}$

Thx very much again, could you please briefly explain how did you get 2/11, how did you work out P(B intersection A) in the formula?
• Apr 27th 2010, 06:18 AM
Plato
Quote:

Originally Posted by dee2020
Thx very much again, could you please briefly explain how did you get 2/11, how did you work out P(B intersection A) in the formula?

Here are the thirty-six outcomes:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
You are told that there is a two in the pair.
So there are only eleven pairs contain a two.
Of those only two pairs multiply to six.
• Apr 27th 2010, 02:09 PM
dee2020
Quote:

Originally Posted by Plato
Here are the thirty-six outcomes:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
You are told that there is a two in the pair.
So there are only eleven pairs contain a two.
Of those only two pairs multiply to six.

thx u soooooo much..
much appreciated :-)