Probability QA 1
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The document provides solutions to various probability questions related to events, including calculations for union, intersection, and individual probabilities. Key problems involve the probabilities of events occurring in relation to students passing examinations in mathematics, biology, and language exams. The document serves as a physics helpline resource, guiding readers through probability concepts and calculations.
![Physics Helpline
L K Satapathy
Q1 : If E and F are events such that
Ans: (i) P(E or F) = P(E F) = P(E) + P(F) – P(E F)
QA Probability - 1
Then find (i) P(E or F) (ii) P(not E and not F)
1 1 1 2 4 1 5 ( )
4 2 8 8 8
ans
(ii) P(not E and not F) = P(E F) = P(E F) [ DeMorgan’s law]
= 1 – P(E F) [ Since P(A) = 1 – P(A) ]
5 31
8 8
P(E) = , P(F) = and P(E and F) =1
4
1
2
1
8](https://image.slidesharecdn.com/qaprobability1-160217132625/85/Probability-QA-1-2-320.jpg)
![Physics Helpline
L K Satapathy
Q4 : In an entrance test which is graded on the basis of two
examinations, the probability of a randomly chosen student passing
the first examination is 0.8 and the probability of passing the second
examination is 0.7. The probability of passing at least one of them is
0.95. What is the probability of passing both ?
Ans : Let A be the event ‘passing in the 1st examination’
QA Probability - 1
and B be the event ‘passing in the 2nd examination’
Given: P(A) = 0.8 , P(B) = 0.7 and P(A B) = 0.95
A B is the event ‘passing both examinations’
and A B is the event ‘passing at least one examination’
Now P(A B) = P(A) + P(B) – P(A B)
P(A B) = P(A) + P(B) – P(A B) = 0.8 + 0.7 – 0.95 = 0.55 [Ans]](https://image.slidesharecdn.com/qaprobability1-160217132625/85/Probability-QA-1-5-320.jpg)
![Physics Helpline
L K Satapathy
Q5 : The probability that a student will pass the final examination in
both English and Hindi is 0.5 and the probability of passing neither is
0.1. If the probability of passing the English examination is 0.75 , what
is the probability of passing the Hindi examination ?
Ans : Let E be the event ‘passing the English examination’
QA Probability - 1
and H be the event ‘passing the Hindi examination’
Given: P(E) = 0.75 , P(H) = ? P(E H) = 0.5 and P(E H) = 0.1
E H is the event ‘passing both examinations’
and (E H) = (E H) is the event ‘passing neither’
P(E H) = 1 – P(E H)= 1 – 0.1 = 0.9
P(E H) = P(E) + P(H) – P(E H)
P(H) = P(E H) + P(E H) – P(E) = 0.9 + 0.5 – 0.75 = 0.65 [Ans]](https://image.slidesharecdn.com/qaprobability1-160217132625/85/Probability-QA-1-6-320.jpg)
Probability QA 1
- 1. Physics Helpline L K Satapathy Probability QA 1
- 2. Physics Helpline L K Satapathy Q1 : If E and F are events such that Ans: (i) P(E or F) = P(E F) = P(E) + P(F) – P(E F) QA Probability - 1 Then find (i) P(E or F) (ii) P(not E and not F) 1 1 1 2 4 1 5 ( ) 4 2 8 8 8 ans (ii) P(not E and not F) = P(E F) = P(E F) [ DeMorgan’s law] = 1 – P(E F) [ Since P(A) = 1 – P(A) ] 5 31 8 8 P(E) = , P(F) = and P(E and F) =1 4 1 2 1 8
- 3. Physics Helpline L K Satapathy Q2 : A and B are events such that P(A) = 0.42 , P(B) = 0.48 Ans: (i) P(not A) = 1 – P(A) = 1 – 0.42 = 0.58 QA Probability - 1 and P(A and B) = 0.16. Determine (ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52 (ii) P(A or B) = P(A) + P(B) – P(A and B) = 0.42 + 0.48 – 0.16 = 0.90 – 0.16 = 0.74 (i) P(not A) (ii) P(not B) and (iii) P(A or B).
- 4. Physics Helpline L K Satapathy Q3 : In class XI of a school, 40% of the students study Mathematics and 30% study Biology. 10% of the students study both Mathematics and Biology. If a student is chosen at random from the class , find the probability that he will be studying Mathematics or Biology. Let M be the event ‘studies Mathematics’ QA Probability - 1 and B be the event ‘studies Biology’ Given: P(M) = 40% = 0.4 , P(B) = 30% = 0.3 and P(M B) = 10% = 0.1 M B is the event ‘studies both Mathematics and Biology’ Ans : For the selected student and M B is the event ‘studies Mathematics or Biology’ P(M B) = P(M) + P(B) – P(M B) = 0.4 + 0.3 – 0.1 = 0.6
- 5. Physics Helpline L K Satapathy Q4 : In an entrance test which is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both ? Ans : Let A be the event ‘passing in the 1st examination’ QA Probability - 1 and B be the event ‘passing in the 2nd examination’ Given: P(A) = 0.8 , P(B) = 0.7 and P(A B) = 0.95 A B is the event ‘passing both examinations’ and A B is the event ‘passing at least one examination’ Now P(A B) = P(A) + P(B) – P(A B) P(A B) = P(A) + P(B) – P(A B) = 0.8 + 0.7 – 0.95 = 0.55 [Ans]
- 6. Physics Helpline L K Satapathy Q5 : The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75 , what is the probability of passing the Hindi examination ? Ans : Let E be the event ‘passing the English examination’ QA Probability - 1 and H be the event ‘passing the Hindi examination’ Given: P(E) = 0.75 , P(H) = ? P(E H) = 0.5 and P(E H) = 0.1 E H is the event ‘passing both examinations’ and (E H) = (E H) is the event ‘passing neither’ P(E H) = 1 – P(E H)= 1 – 0.1 = 0.9 P(E H) = P(E) + P(H) – P(E H) P(H) = P(E H) + P(E H) – P(E) = 0.9 + 0.5 – 0.75 = 0.65 [Ans]
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