Probability QA 3
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The document presents solutions to probability problems involving drawing colored balls, mutually exclusive events, and the sum of numbers from thrown dice. Specific examples include calculating the probability of drawing two balls of different colors from a mixed bag, determining probabilities of mutually exclusive and exhaustive events, and finding the probability of the sum of two dice being neither divisible by 3 nor by 4. Detailed calculations and definitions are provided for each scenario.
![Physics Helpline
L K Satapathy
QA Probability - 3
2 white balls can be drawn from 2 white balls in ways2
2C
2 red balls can be drawn from 3 red balls in ways3
2C
2 green balls can be drawn from 5 green balls in ways5
2C
2 black balls can be drawn from 4 black balls in ways4
2C
2 3 5 4
2 2 2 2( )n E C C C C
3 2 5 4 4 31 1 3 10 6 20
1 2 1 2 1 2
( ) 20( )
( ) 91
n E
P E
n S
20 71( ) 1 ( ) 1
91
[ ]
91
P E P E Ans ](https://image.slidesharecdn.com/qaprobability3-160220114113/85/Probability-QA-3-3-320.jpg)
![Physics Helpline
L K Satapathy
Q2 : If A , B and C are three mutually exclusive and exhaustive events
associated with a random experiment , then find P(A).
QA Probability - 2
Ans :
A , B and C are mutually exclusive A B = B C = A C =
It is given that P(B) = P(A) and P(C) = P(B)3
2
1
2
and exhaustive A B C = S
P(ABC) = P(A) + P(B) + P(C) = P(S) = 1 . . . (1)
3 3 1 3( ) ( ) ( ) ( ) ( )
2 2 2 4
Let P A p P B P A p and P C P B p
3 3 4 6 3(1) 1 1
2 4 4
p p p p
4( )
13
[ ]AnsP A p ](https://image.slidesharecdn.com/qaprobability3-160220114113/85/Probability-QA-3-4-320.jpg)
![Physics Helpline
L K Satapathy
QA Probability - 3
( ) 12 1( )
( ) 36 3
n A
P A
n S
( ) 9 1( )
( ) 36 4
n B
P B
n S
( ) 1( )
( ) 36
n A B
P A B
n S
( ) ( ) ( ) ( )P A B P A P B P A B
1 1 1 12 9 1 20 5
3 4 36 36 36 9
5 4( ) 1 ( ) 1
9 9
[ ]P A B P A B Ans ](https://image.slidesharecdn.com/qaprobability3-160220114113/85/Probability-QA-3-6-320.jpg)
Probability QA 3
- 1. Physics Helpline L K Satapathy Probability QA 3
- 2. Physics Helpline L K Satapathy Q1 : Two balls are drawn at random from a bag containing 2 white , 3 red , 5 green and 4 black balls , one by one , without replacement. Find the probability that both the balls are of different colors. QA Probability - 3 Ans : There are 2 white 3 red , 5 green and 4 black balls Total number of balls = 2 + 3 + 5 + 4 = 14 2 balls can be drawn from 14 balls in ways14 2C 14 2 14 13( ) 91 1 2 n S C Let the event ‘both are of same color’ = E the event ‘both are of different colors’ = E E = Both are of the same color both white , both red , both green or both black P(E) = 1 – P(E)
- 3. Physics Helpline L K Satapathy QA Probability - 3 2 white balls can be drawn from 2 white balls in ways2 2C 2 red balls can be drawn from 3 red balls in ways3 2C 2 green balls can be drawn from 5 green balls in ways5 2C 2 black balls can be drawn from 4 black balls in ways4 2C 2 3 5 4 2 2 2 2( )n E C C C C 3 2 5 4 4 31 1 3 10 6 20 1 2 1 2 1 2 ( ) 20( ) ( ) 91 n E P E n S 20 71( ) 1 ( ) 1 91 [ ] 91 P E P E Ans
- 4. Physics Helpline L K Satapathy Q2 : If A , B and C are three mutually exclusive and exhaustive events associated with a random experiment , then find P(A). QA Probability - 2 Ans : A , B and C are mutually exclusive A B = B C = A C = It is given that P(B) = P(A) and P(C) = P(B)3 2 1 2 and exhaustive A B C = S P(ABC) = P(A) + P(B) + P(C) = P(S) = 1 . . . (1) 3 3 1 3( ) ( ) ( ) ( ) ( ) 2 2 2 4 Let P A p P B P A p and P C P B p 3 3 4 6 3(1) 1 1 2 4 4 p p p p 4( ) 13 [ ]AnsP A p
- 5. Physics Helpline L K Satapathy Q3 : Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 4? QA Probability - 3 Ans : Random experiment : Throwing of 2 dice. n(S) = 36 Let the event ‘sum divisible by 3’ = A And the event ‘sum divisible by 4’ = B We are required to find P(A B) = P(A B) = 1 – P(A B) Now A = {12 , 15 , 21 , 24 , 33 , 36 , 42 , 45 , 51 , 54 , 63 , 66} n(A) = 12 And B = {13 , 22 , 26 , 31 , 35 , 44 , 53 , 62 , 66} n(B) = 9 A B = { 66 } n(A B) = 1
- 6. Physics Helpline L K Satapathy QA Probability - 3 ( ) 12 1( ) ( ) 36 3 n A P A n S ( ) 9 1( ) ( ) 36 4 n B P B n S ( ) 1( ) ( ) 36 n A B P A B n S ( ) ( ) ( ) ( )P A B P A P B P A B 1 1 1 12 9 1 20 5 3 4 36 36 36 9 5 4( ) 1 ( ) 1 9 9 [ ]P A B P A B Ans
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