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Probablity Question

by Utkarsh86 » Mon Feb 28, 2011 7:40 pm
Suppose that 2 dices are tossed. For each dice it is equally likely that 1,2,3,4,5 or 6 dots will show.
"¢ What is the probability that the sum of the dots on the uppermost faces of the two dice will be 5 or 7 ?

"¢ What is the prob that the sum of the dots on the uppermost faces of the two dice will be some other than 4 or 8.

"¢ Let El be the event that the first die shows a 3. Let E2 be the even that the sum of the dots on the uppermost faces of the two dice is 6. Are E1 and E 2 independent events?

"¢ Again let E1 be the even that the first die shows a 3. Let E3 be the event that the sum of the dots on the uppermost faces of the dice is 7. Are E1 and E3 independent events

"¢ Given that the sum of the dots on the uppermost faces of the two dice is 7, what is the probability that the first dice is 7, what is the probability that the first die shows 4 dots ?

"¢ Given that the first die shows a 3, what is the probability that the sum of the dots on the uppermost faces of the 2 dice is an even number


Ive been trying to solve it but i get stuck all the time all help is appreciated
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by Anurag@Gurome » Mon Feb 28, 2011 8:59 pm
Suppose that 2 dices are tossed. For each dice it is equally likely that 1,2,3,4,5 or 6 dots will show.
"¢ What is the probability that the sum of the dots on the uppermost faces of the two dice will be 5 or 7 ?

Required probability is (Number of favorable cases)/(Total number of all possible cases).
Total number of all possible cases is 6*6 = 36.
Favorable cases are (1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).
Their total number is 10.
Or required probability is 10/36 = 5/18.
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by manpsingh87 » Mon Feb 28, 2011 9:36 pm
Suppose that 2 dices are tossed. For each dice it is equally likely that 1,2,3,4,5 or 6 dots will show.

A)What is the probability that the sum of the dots on the uppermost faces of the two dice will be 5 or 7 ?

B)What is the prob that the sum of the dots on the uppermost faces of the two dice will be some other than 4 or 8.



As two dices are thrown therefore total number of outcomes would be 6^2 =36;

A) sum will be 5 or 7: 5 will occur in {(1,4)(2,3)(3,2)(4,1)} and 7 will occur in {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)} 6 ways therefore total number of favorable outcomes are 10.
Hence the desired probability is 10/36 = 5/18

B) Sum will be 4 or 8: 4 will occur in {(1,3)(2,2)(3,1)} 3 ways; 8 will occur in {(2,6)(3,5)(4,4)(5,3)(6,2)} 5 ways

P(occurrence of an event) = 1 - P( Non-occurrence of an event)
P(other than 4 or 8) = 1 - P(4 or 8)
= 1- (8/36) = 1 - (2/9) = 7/9
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by Anurag@Gurome » Mon Feb 28, 2011 9:41 pm
What is the prob that the sum of the dots on the uppermost faces of the two dice will be some other than 4 or 8.


Solution:
Let us calculate the probability that the sum is 4 or 8.
The favorable cases are (1, 3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4).
Their number is 8.
Total number of all possible cases is 6*6 = 36.
Or, required probability is 8/36 = 2/9.
So, the probability that the sum is something other than 4 or 8 is 1 - 2/9 = 7/9.
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by Utkarsh86 » Mon Feb 28, 2011 10:02 pm
"¢ Given that the sum of the dots on the uppermost faces of the two dice is 7, what is the probability that the first die shows 4 dots ?

"¢ Given that the first die shows a 3, what is the probability that the sum of the dots on the uppermost faces of the 2 dice is an even number


The help above was useful i figured out but i am a little confused here.

There are 6 ways to get a 7, but there is 1/6 ways to get a 4 on a dice and 1/6 ways to get a 4 on the 2nd dice, this is the part im getting confused. How do i calculate it ?
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by Anurag@Gurome » Mon Feb 28, 2011 11:27 pm
Given that the sum of the dots on the uppermost faces of the two dice is 7, what is the probability that the first dice shows 4 dots ?

This is a question on conditional probability.
So, P(AlB) = P(A and B)/P(B)
B is the event that that sum of dots on faces is 7.
The favorable cases are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).
P(B) = 6/36 = 1/6.
A is the event that first dice shows 4.
So (A and B) is the case (4, 3).
Or P(A and B) is 1/36.
So, P(AlB) is (1/36)/(1/6) = 1/6.
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by manpsingh87 » Mon Feb 28, 2011 11:43 pm
Anurag@Gurome wrote:Given that the sum of the dots on the uppermost faces of the two dice is 7, what is the probability that the first dice shows 4 dots ?

This is a question on conditional probability.
So, P(AlB) = P(A and B)/P(B)
B is the event that that sum of dots on faces is 7.
The favorable cases are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).
P(B) = 6/36 = 1/6.
A is the event that first dice shows 4.
So (A and B) is the case (4, 3).
Or P(A and B) is 1/36.
So, P(AlB) is (1/36)/(1/6) = 1/6.
hi sir, i think as the no. of ways of getting a sum of 7 is 6 {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} and out these only one is a desired one i.e. (4,3) so i believe we can also safely conclude that probability will be 1/6..!! please correct me if i'm wrong..!!!!!
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by Anurag@Gurome » Tue Mar 01, 2011 12:04 am
manpsingh87 wrote:
Anurag@Gurome wrote:Given that the sum of the dots on the uppermost faces of the two dice is 7, what is the probability that the first dice shows 4 dots ?

This is a question on conditional probability.
So, P(AlB) = P(A and B)/P(B)
B is the event that that sum of dots on faces is 7.
The favorable cases are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).
P(B) = 6/36 = 1/6.
A is the event that first dice shows 4.
So (A and B) is the case (4, 3).
Or P(A and B) is 1/36.
So, P(AlB) is (1/36)/(1/6) = 1/6.
hi sir, i think as the no. of ways of getting a sum of 7 is 6 {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} and out these only one is a desired one i.e. (4,3) so i believe we can also safely conclude that probability will be 1/6..!! please correct me if i'm wrong..!!!!!
Here, for this problem, your logic is perfectly correct.
But, generally it is advisable to use the formula to avoid errors.

Suppose the question is: what is the probability, that the sum on two faces is greater than 9, given that first dice shows 5?
How will you solve this?
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by manpsingh87 » Tue Mar 01, 2011 1:05 am
Anurag@Gurome wrote:
manpsingh87 wrote:
Anurag@Gurome wrote:Given that the sum of the dots on the uppermost faces of the two dice is 7, what is the probability that the first dice shows 4 dots ?

This is a question on conditional probability.
So, P(AlB) = P(A and B)/P(B)
B is the event that that sum of dots on faces is 7.
The favorable cases are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).
P(B) = 6/36 = 1/6.
A is the event that first dice shows 4.
So (A and B) is the case (4, 3).
Or P(A and B) is 1/36.
So, P(AlB) is (1/36)/(1/6) = 1/6.
hi sir, i think as the no. of ways of getting a sum of 7 is 6 {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} and out these only one is a desired one i.e. (4,3) so i believe we can also safely conclude that probability will be 1/6..!! please correct me if i'm wrong..!!!!!
Here, for this problem, your logic is perfectly correct.
But, generally it is advisable to use the formula to avoid errors.

Suppose the question is: what is the probability, that the sum on two faces is greater than 9, given that first dice shows 5?
How will you solve this?
sum greater than 9, i.e. sum can be either 10,11,12, now these sums can be achieved in {(4,6)(6,4)(5,5)(5,6)(6,5)(6,6)} 6 ways therefore the desired probability as per the traditional method suggested by anurag sir above would be

(2/36)/(6/36) = 1/3;

and sir as per my logic it would be 2/6 as the two favorable cases would be (5,6)(5,5) =1/3;

sir still getting the same answer by both the methods...!!!!
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by Anurag@Gurome » Tue Mar 01, 2011 7:29 pm
manpsingh87 wrote:
Anurag@Gurome wrote:
manpsingh87 wrote:
Anurag@Gurome wrote:Given that the sum of the dots on the uppermost faces of the two dice is 7, what is the probability that the first dice shows 4 dots ?

This is a question on conditional probability.
So, P(AlB) = P(A and B)/P(B)
B is the event that that sum of dots on faces is 7.
The favorable cases are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3).
P(B) = 6/36 = 1/6.
A is the event that first dice shows 4.
So (A and B) is the case (4, 3).
Or P(A and B) is 1/36.
So, P(AlB) is (1/36)/(1/6) = 1/6.
hi sir, i think as the no. of ways of getting a sum of 7 is 6 {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} and out these only one is a desired one i.e. (4,3) so i believe we can also safely conclude that probability will be 1/6..!! please correct me if i'm wrong..!!!!!
Here, for this problem, your logic is perfectly correct.
But, generally it is advisable to use the formula to avoid errors.

Suppose the question is: what is the probability, that the sum on two faces is greater than 9, given that first dice shows 5?
How will you solve this?
sum greater than 9, i.e. sum can be either 10,11,12, now these sums can be achieved in {(4,6)(6,4)(5,5)(5,6)(6,5)(6,6)} 6 ways therefore the desired probability as per the traditional method suggested by anurag sir above would be

(2/36)/(6/36) = 1/3;

and sir as per my logic it would be 2/6 as the two favorable cases would be (5,6)(5,5) =1/3;

sir still getting the same answer by both the methods...!!!!
Yes, the answer is 1/3.
I think, if you are confident in your approach, you can always use the method which suits you!
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