Probability QA 12
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The document discusses a probability question involving three tosses of a fair die and the assessment of a specific sum related to complex cube roots of unity. It details calculations for the total possible outcomes and how to find the probability that the sum equals zero. The correct option is noted along with relevant combinations and arrangements of the die results.
![Physics Helpline
L K Satapathy
Probability QA 12
Correct option = (c)
Now r1 , r2 and r3 are of the form 3n , 3n+1 , 3n+2
Let r1 take the values (3 or 6)
For each of these r2 takes the values (1 or 4) and r3 takes the values (2 or 5)
n(E) = 2226 = 48 . . . (2)
( ) 48 2(1) (2) ( )
( ) 216
[ ]
9
n E
P nE
S
A s
n
We are required to find the probability that the sum = zero
1 2 3
0
r r r
The number of cases when the sum is zero = 222 = 8
Also the values of r1 , r2 and r3 can be rearranged in 3! = 6 ways
The Total number of cases when the sum is zero = 2226](https://image.slidesharecdn.com/qaprobability12-161104125756/85/Probability-QA-12-3-320.jpg)
Probability QA 12
- 1. Physics Helpline L K Satapathy Probability QA 12
- 2. Physics Helpline L K Satapathy Probability QA 12 Question : Let be a complex cube root of unity with 1. A fair die is thrown three times. If r1 , r2 and r3 are the numbers obtained on the die , then the probability that is Answer : 1 2 3 0 r r r 1 1 2 1( ) ( ) ( ) ( ) 18 9 9 36 a b c d We have r1 , r2 and r3 are the number s obtained in the three tosses It may be noted that the values of the sum can be zero or non-zero Each of r1 , r2 and r3 can take six values Each term of the sum can take six values from to1 6 1 2 3r r r n(S) = 666 = 216 . . . (1)
- 3. Physics Helpline L K Satapathy Probability QA 12 Correct option = (c) Now r1 , r2 and r3 are of the form 3n , 3n+1 , 3n+2 Let r1 take the values (3 or 6) For each of these r2 takes the values (1 or 4) and r3 takes the values (2 or 5) n(E) = 2226 = 48 . . . (2) ( ) 48 2(1) (2) ( ) ( ) 216 [ ] 9 n E P nE S A s n We are required to find the probability that the sum = zero 1 2 3 0 r r r The number of cases when the sum is zero = 222 = 8 Also the values of r1 , r2 and r3 can be rearranged in 3! = 6 ways The Total number of cases when the sum is zero = 2226
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