1. ## probability question

roll 2 dice. Find the probability that:

a)the sum is atleast 9
b)the sum is either less than 4 or greater than 9
c)the sum is atleast 10, given that the sum is at least 9.
d) the sum is more than 7, given that the sum is no more than 5.

thank you very much for helping.

2. Originally Posted by shellshock
roll 2 dice. Find the probability that:

a)the sum is atleast 9
b)the sum is either less than 4 or greater than 9
c)the sum is atleast 10, given that the sum is at least 9.
d) the sum is more than 7, given that the sum is no more than 5.

thank you very much for helping.
Use the following dice table: Dice table

If you need more help, please show what you've been able to do and state precisely where you are stuck.

3. ## Hello !

Originally Posted by shellshock
roll 2 dice. Find the probability that:

a)the sum is atleast 9
b)the sum is either less than 4 or greater than 9
c)the sum is atleast 10, given that the sum is at least 9.
d) the sum is more than 7, given that the sum is no more than 5.

thank you very much for helping.

First what u need to do is make a sample table of the 2 die's
like this

(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)

Let the sample space be denoted as S = 36

Now
a)Probability of sum being atleast 9 ie Pr(X $\le$ 9)

which is
Pr(X $\le$ 9) = Pr(X = 9) + Pr(X = 10) + Pr(X = 11) + Pr(X = 12)

= $\frac{4}{36} + \frac{3}{36} + \frac{2}{36} + \frac{1}{36}$ = .....rest i will leave up to u to compute

b)Probability of sum less than 4 or greater than 9

Take Pr(A) = Probability of sum being less than 4 ie(1,1) (1,2) (2,1)

Pr(A) = $\frac{3}{36}$ = $\frac{1}{12}$

Take Pr(B) = Probability of sum being greater than 9 ie. (4,6) (5,5) (6,4) (5,6) (6,5) (6,6)

$\therefore$ Pr(B) = $\frac{6}{36} = \frac{1}{6}$

Now question is prob of sum less than 4 or greater than 9 and here or means $\cup$

$\therefore$ Pr(A $\cup$ B) = Pr(A) + Pr(B) .......since they both are mutually exclusive

= $\frac{1}{12} + \frac{1}{6}$ = .............and u can do it from here on by urself

c)Probability that the sum is atleast 10, given that the sum is atleast 9

here Pr(A) = sum atleast 10...........which we need to find out

Pr(B) = given that the sum is atleast 9

Pr(A $\cap$ B) = reading which are both in Pr(X $\le$ 9) and when Pr(X $\le$ 10) ie (4,5) (5,5) (5,4) (6,5) (5,6) (6,6) = .........
$\therefore$ $Pr(A \cap B)$ = $\frac{1}{6}$

there using conditional prob

Pr(A|B) = $\frac{Pr(A \cap B)}{Pr(B)}$ = ...........substitute and get the answer

d) the sum is more than 7, given that the sum is no more than 5.

It is also quite similar to the c) .........so u should be able to do it

4. thank you very much, I have figured it out.

5. Originally Posted by zorro
First what u need to do is make a sample table of the 2 die's
like this

(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)

Let the sample space be denoted as S = 36

Now
a)Probability of sum being atleast 9 ie Pr(X $\le$ 9)

which is
Pr(X $\le$ 9) = Pr(X = 9) + Pr(X = 10) + Pr(X = 11) + Pr(X = 12) Mr F says: This is ${\color{red}Pr(X \geq 9)}$
. No harm done though, the OP just needs to subtract from 1 the result of the calculation below.

= $\frac{4}{36} + \frac{3}{36} + \frac{2}{36} + \frac{1}{36}$ = .....rest i will leave up to u to compute

b)Probability of sum less than 4 or greater than 9

Take Pr(A) = Probability of sum being less than 4 ie(1,1) (1,2) (2,1)

Pr(A) = $\frac{3}{36}$ = $\frac{1}{12}$

Take Pr(B) = Probability of sum being greater than 9 ie. (4,6) (5,5) (6,4) (5,6) (6,5) (6,6)

$\therefore$ Pr(B) = $\frac{6}{36} = \frac{1}{6}$

Now question is prob of sum less than 4 or greater than 9 and here or means $\cup$

$\therefore$ Pr(A $\cup$ B) = Pr(A) + Pr(B) .......since they both are mutually exclusive

= $\frac{1}{12} + \frac{1}{6}$ = .............and u can do it from here on by urself

c)Probability that the sum is atleast 10, given that the sum is atleast 9

here Pr(A) = sum atleast 10...........which we need to find out

Pr(B) = given that the sum is atleast 9

Pr(A $\cap$ B) = reading which are both in Pr(X $\le$ 9) and when Pr(X $\le$ 10) ie (4,5) (5,5) (5,4) (6,5) (5,6) (6,6) = .........
$\therefore$ $Pr(A \cap B)$ = $\frac{1}{6}$

there using conditional prob

Pr(A|B) = $\frac{Pr(A \cap B)}{Pr(B)}$ = ...........substitute and get the answer

d) the sum is more than 7, given that the sum is no more than 5.

It is also quite similar to the c) .........so u should be able to do it
..