# Math Help - Probability/Two fair dice are tossed

1. ## Probability/Two fair dice are tossed

Two fair dice are tossed and the product of the outcomes is recorded:

(a)Write down the sample space s, the mode and median of S

S={1,2,3,4,5,6,2,4,6,8,10,12,3,6,9,12,15,18,4,8,12 ,16,20,24,5,10,15,20,25,30,6,12,18,24,30,36}

Mode: 6 and 12 (Bimodal)

Medain=10

(b)compute the probabilities of the events below:

(i)Product=9

P(9)=P(3*3)=1/36

(ii)Product=13

P(13)=0

Can I write P(13)=Ф

(iii)Product is a perfect square greater than 2

which of my answers is true :

P(perfect square >2)=P{(2,2) ,(3,3),(4,4)(5,5),(6,6)}=5/36

or

p(perfect square >2)=p(4)+p(9)+P(16)+P(25)

=3/36 +1/36 +1/36 +1/36 +1/36=7/36

(iv)Product=12

p(12)=p{(3,4),(4,3),(2,6),(6,2)}=4/36

(c)Compute the expected value for the product

What is the meaning of expected value?and How can I obtain it?

(d)Compute the probabilities of obtaining a double six and the expected number of attempts till a double six is attained

Does the question mean Probability of obtaining a double 6, (6,6) is 1/36.?

and how can i obtain the expected number of attempts till a double 6 is attained

2. Originally Posted by change_for_better
Two fair dice are tossed and the product of the outcomes is recorded:

(a)Write down the sample space s, the mode and median of S

S={1,2,3,4,5,6,2,4,6,8,10,12,3,6,9,12,15,18,4,8,12 ,16,20,24,5,10,15,20,25,30,6,12,18,24,30,36}

Mode: 6 and 12 (Bimodal)

Medain=10

(b)compute the probabilities of the events below:

(i)Product=9

P(9)=P(3*3)=1/36

(ii)Product=13

P(13)=0

I agree with all of these

Originally Posted by change_for_better

(iii)Product is a perfect square greater than 2

which of my answers is true :

p(perfect square >2)=p(4)+p(9)+P(16)+P(25)

=3/36 +1/36 +1/36 +1/36 +1/36=7/36

(iv)Product=12

p(12)=p{(3,4),(4,3),(2,6),(6,2)}=4/36
These answers I also agree with

Originally Posted by change_for_better

(c)Compute the expected value for the product

(d)Compute the probabilities of obtaining a double six and the expected number of attempts till a double six is attained
c)

$E(X) = 1\times p(1)+2\times p(2)+ \cdots + 36 \times p(36)$

$E(X) = 1\times \frac{1}{36}+2\times +\frac{2}{36} \cdots + 36 \times \frac{1}{36}$

d) $\mu = n\times p$ where $\mu = E(X)$ from the previous problem and $p = \frac{1}{36}$

so $n = \frac{\mu}{p} = \frac{E(X)}{p}$

3. Originally Posted by pickslides
I agree with all of these

These answers I also agree with

c)

$E(X) = 1\times p(1)+2\times p(2)+ \cdots + 36 \times p(36)$

$E(X) = 1\times \frac{1}{36}+2\times +\frac{2}{36} \cdots + 36 \times \frac{1}{36}$

d) $\mu = n\times p$ where $\mu = E(X)$ from the previous problem and $p = \frac{1}{36}$

so $n = \frac{\mu}{p} = \frac{E(X)}{p}$

Thank you very much ..

Can you explain the last point:

d) $\mu = n\times p$ where $\mu = E(X)$ from the previous problem and $p = \frac{1}{36}$

so $n = \frac{\mu}{p} = \frac{E(X)}{p}$[/quote]