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What is the required probability?When two dice are thrown what is the probability that...
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High School Teacher
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The numbers on the dice for which the sum of the numbers can be 6 is: (1,5) , (2, 4), (3,3) , (4,2), (5,1).
Now we need to get one event out of six possible events. So the probability of getting any particular number when a die is tossed is 1/6.
Also, the probability of two events occurring simultaneously is the product of the probabilities of their occurring individually.
Therefore the probability of each of the five cases we have to consider occuring is:
The first die has 1 and the second 5: the probability is 1/6 * 1/6 = 1/36
The first die has 2 and the second 4: the probability is 1/6 * 1/6 = 1/36
The first die has 3 and the second 3: the probability is 1/6 * 1/6 = 1/36
The first die has 4 and the second 2: the probability is 1/6 * 1/6 = 1/36
The first die has 5 and the second 1: the probability is 1/6 * 1/6 = 1/36
To get the sum as 6 any of these cases will do. So the probability that we get a 6 is the sum of the individual probabilities we just derived, which is equal to 5*(1/36)= 5/36.
Therefore the probability that we get the sum as 6 when two dice are thrown is 5/36.
Posted by sociality on October 27, 2010 at 4:12 AM (Answer #1)
To solve this question we will consider one dice at time.
The first dice mus have any number between 1 to 5. If the dice is has number 6 than the sum of the two dice will definitely be greater than 6.
Once the first dice has any number from 1 to 5, the second dice myst gat a specific number so that the sum of the number on two dice is 6. Thus if the first dice has a number n, then the second dice must have a specific matching number given by
Number on second dice = 6 - n
The probability on of getting sum of 6 for the two dice can now be calculated as:
Probability of getting a sum of 6 for the two dice:
= (Probability of getting any number 'n' from 1 to 5 on first dice) x (Probability of getting a specific matching number equal to 6-n on second dice)
= (5/6) x (1/6)
Posted by krishna-agrawala on October 27, 2010 at 3:45 PM (Answer #2)
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