Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 10)
10.
Two dice are tossed. The probability that the total score is a prime number is:
Answer: Option
Explanation:
Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
| Then E | = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) } |
n(E) = 15.
| P(E) = |
n(E) | = | 15 | = | 5 | . |
| n(S) | 36 | 12 |
Discussion:
64 comments Page 5 of 7.
Anil singh said:
1 decade ago
Prime number divisible by 1 and that nimber only but in the case (1,4), (1,6) etc divided by other number like 1,2,3
Please explain.
Please explain.
Rakesh said:
1 decade ago
Why we not chosen (2, 2) because it also prime number and it divisible by only 2?
Ramya said:
1 decade ago
They have left out 6, 3 and 3, 6 then 5, 6 and 6, 5. These also should be included then the total probability becomes 1/2.
Ravindra said:
1 decade ago
@Harshini.
Permutations are used for arrangement where as combinations are used for selection.
Ex: Arrange the name xyz then per is used.
Where as,
Select one among this xyz comb is used.
Permutations are used for arrangement where as combinations are used for selection.
Ex: Arrange the name xyz then per is used.
Where as,
Select one among this xyz comb is used.
Siva said:
1 decade ago
Prime number means the we gets only one time. i.e., 1*2=2 only once we get 2. But for 4 1*4=4, 2*2=4 now we get 4 for 2 times in-terms of multiplication.
Ganesh Sang said:
1 decade ago
Please in every example we need to find samples or experiment samples or there is any shortcut for easily getting number of prime number or sum of two numbers as a prime ?
Suvendu said:
1 decade ago
As the question says sum should be a prime number.
So sum must be 2, 3, 5, 7, 11,
So probability should be (1, 1) (1, 2) (2, 1) (1, 4) (4, 1) (3, 4) (4, 3) (5, 6) (6, 5).
So probability= (9/36) =1/4.
So sum must be 2, 3, 5, 7, 11,
So probability should be (1, 1) (1, 2) (2, 1) (1, 4) (4, 1) (3, 4) (4, 3) (5, 6) (6, 5).
So probability= (9/36) =1/4.
Akshay said:
1 decade ago
Why don't we consider (1, 1) two times. Because it will be the outcome of both the dices.
Sowmya said:
1 decade ago
The sample space is a set, so repeated event (1, 1) is taken only once.
Ram reddy said:
1 decade ago
Please tell me any other method.
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