5. Addition Theorem of Probability
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For two events
P(AUB) = P(A) + P(B)- P(A∩B)
For Independent events, P(A∩B) = P(A) X P(B)
For mutually exclusive events, P(A∩B) = 0
For three events
P (Either A or B or C) = P (A) + P (B) + P (C) – P (AB) – P (AC) – P (BC) + P
(ABC)
6. Addition Theorem of Probability
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A card is drawn at random from a well shuffled pack of
cards, find the probability of the card being a diamond or a
queen
Answer : 4/13
7. Conditional Probability - Theory
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A and B are two events associated with a random experiment.
Probability of occurrence of B, given that A has already occurred
P (B/A) =
𝑃 (𝐴 𝑎𝑛𝑑 𝐵)
𝑃 (𝐴)
8. Conditional Probability
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A bag contains 10 white and 5 black balls. Two balls are
drawn at random one after the other without replacement.
Find the probability that both balls drawn are black.
Answer : 2/21
9. Conditional Probability
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Find the probability of drawing a king, a queen and a
knave in that order from a pack of cards in three
consecutive draws, the cards drawn not being
replaced.
Answer : 4/52 x 4/51 x 4/50
10. Conditional Probability
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A speaks truth in 80% cases, B in 90% cases. In
what percentage of cases are they likely to contradict
a each other in answering a yes/no question?
Answer : 13/50 = 26%
11. Law of Total Probability - Theory
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Suppose we have a set of mutually exclusive and collectively exhaustive
events, B1, B2, . . . Bk , then any other event A is a union of its pieces:
A = (A ∩ B1) ∪ (A ∩ B2) ∪ . . . ∪ (A ∩ Bk )
Those pieces are disjoint, so
P(A) = P(A ∩ B1) + P(A ∩ B2) + . . . + P(A ∩ Bk )
Applying the multiplication to the above, we get:
P(A) = P(A|B1)P(B1) + P(A|B2)P(B2) + . . . + P(A|Bk )P(Bk )
12. Law of Total Probability
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Suppose we have two hats: one has 4 red balls and 6 green balls,
the other has 6 red and 4 green. We toss a fair coin, if heads, pick a
random ball from the first hat, if tails from the second. What is the
probability of getting a red ball?
Answer : 1/2
13. Law of Total Probability
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A soccer team wins 60% of its games when it scores the first goal,
and 10% of its games when the opposing team scores first. If the
team scores the first goal about 30% of the time, what fraction of
the games does it win?
Answer : 1/4
14. Bayes’ theorem
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There are three boxes, B1, B2 and B3. B1 contains 5 blue
and 6 black balls. B2 contains 4 blue and 5 red balls. B3
contains 3 blue and 4 green balls. If one of the boxes is
selected at random and a ball is drawn from it and found to
be blue, then what is the probability that it was drawn from
B2?
Answer:
5
9
5
11
+
5
9
+
3
7
15. BERNOULLI’S THEOREM - Theorem
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If the probability of happening of an event on one trial or experiment is known,
then the probability of its happening exactly, 1,2,3,…r times in n trials can be
determined by using the formula:
P (r) = nCr pr . qn-r r = 1,2,3,…n
Where,
P (r) = Probability of r successes in n trials.
p = Probability of success or happening of an event in one trial.
q = Probability of failure or not happening of the event in one trial.
n = Total number of trials.
17. BERNOULLI’S THEOREM
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A test consists of 10 multiple choice questions with five
choices for each question. As an experiment, you GUESS
on each and every answer without even reading the
questions.
What is the probability of getting exactly 6 questions
correct on this test?
Answer: 10𝐶6 ∗
1
5
6
∗
4
5
4
18. Practice Problems
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A child takes a step forward with a probability of 0.3 and
backward with a probability of 0.7. what is the probability
that at the end of 13 steps, the child is one step away from
the starting point?
1. 13C7* (0.21)5
2. 13C7* (0.21)7
3. 13C7* (0.21)6
4. None of these
Answer: Option 3
19. Practice Problems
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A man speaks truth 4 out of 7 times. He throws a die and
reports that it is a five. Find the probability that it is
actually 5.
1. 17/42
2. 19/42
3. 23/42
4. None of these
Answer: 4/19
20. Practice Problems
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What is the probability that an ordinary year has 53
Sundays?
Ans : 1/7
What is the probability that a leap year has 53 Sundays?
Ans : 2/7
What is the probability that a leap year has 53 Sundays
and 53 Mondays?
Ans: 1/7
22. Practice Problems
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Three vertices of a regular hexagon are chosen at random.
What is the probability that the triangle formed with these
three vertices is not an equilateral triangle?
Ans: 0.9
23. Practice Problems
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If two square from a regular 8*8 chessboard are chosen at
random, what is the probability that the two squares don’t
have any side in common
1. 17/19
2. 17/18
3. 2/18
4. 11/18
5. None of these
Answer: 17/18
Editor's Notes
Die shows 5 and he tells 5 = 1/6 *4/7
Die shows other than 5 and he tells 5 = 5/6 * 3/7
Sum of the two values in the two cases.
Die shows 5 and he tells 5 = 1/6 *4/7
Die shows other than 5 and he tells 5 = 5/6 * 3/7
Sum of the two values in the two cases.
https://www.quora.com/Whats-the-probability-that-a-leap-year-has-53-Sundays
So the probability of 53 Sundays in a leap year is 2/7.
A non-leap year has 365 days or 52 weeks and 1 odd day. The odd day can be Sunday,Monday, Tuesday,Wednesday,Thursday,Friday or Saturday.
So there are 7 possibilities out of which 1 is favorable. So the probability of 53 Sundays in non-leap year is 1/7.So the probability of 53 Sundays in a year is 1/4×2/7+3/4×1/7=2/28+3/28=5/28
