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High School Mathematics - 2
13.6 Conditional Probability

Conditional Probability

Suppose a card is removed from a pack of cards and kept aside.
Let B be the event that the card is a spade.
After this one card is drawn at random from the remaining 51 cards.
Let A be the event that the second card is a spade. Under the assumption that B has happened n(A) = 12
P(A/B) = 12/51
It is read as probability of A when B has taken place or probability of A under B.
P(A/B) = n(A B ) / n(B)

Multiplication Theorem of Probability

    If A and B are any 2 events such that
  • P(A) ≠ 0
    then P(A B ) = P(A) . P(B/A)
  • P(B) ≠ 0
    then P(A B) = P(B).P(A/B)
  • P(B/A) = P(B)( i.e. Independent events)
    then P(A/B) = P(A)
Solved Examples
  1. A card is drawn from a pack of cards . Given that it is a heart, what is the probability that it is not a face card?

    Solution
    A = The card is a heart.
    B = It is not a face card.
    As there are 3 face cards of heart, n(A B ) = 13-3 = 10.
    n(A) = 13
    So the required probability is
    P(B/A) = n(A B )/n(A) = 10/13

  2. The probability that a student A can solve a problem is 2/3 and B can solve it 5/8 and C can solve it is 1/4. If all of them try it independently, what is the probability that the problem is solved.

    Solution
    Given P(A) = 2/3, P(B) = 5/8, P(C) = 1/4
    So P() = 1/3, P(B') = 3/8, P(C') = 3/4.
    Consider the event that the problem is not solved. i.e. B' C';
    So the required probability is
    P(problem solved) = 1 - P(problem not solved)
    = 1 - P( B' C';)
    = 1 - P().P(B').P(C')
    = 1 - (1/3.3/8.3/4)
    = 29/32

  3. One bag contains 4 apples and 3 mangoes and another contains 3 apples and 6 mangoes. One fruit is transferred at random from the first bag to the second and then a fruit is selected from the second bag. Find the probability that the fruit selected is an apple.

    Solution
    Let A be the event that an apple is transferred from the first bag and let B be the event that an apple is selected from the second bag.
    So B may happen with A or
    So B = (A B ) U (A' B)
    A B and B are mutually exclusive.
    So P(B) = P(A B) + P( B).
    P( A B) = P(A) . P(B/A)
    = 4/7.4/10= 16/70
    P( B) = P() . P( B/)
    = 3/7.3/10
    So P(B) = P( A B) + P( B)
    = 16/70 + 9/70 = 25/70 = 5/14
    So P(B) = 5/14


Q 1: If the chance of A winning a race is 1/6 and the chance of B winning it is 1/8, what is the chance that neither should win?
12/19
18/37
35/48
7/23

Q 2: A box contains 25 tickets numbered 1-25. One of the ticket is drawn at random and kept aside. Then a second ticket is drawn. Find the probability that both tickets show a number greater than 10.
12/39
7/39
11/20
7/20

Q 3: A box contains 10 tickets numbered from 1- 10. 2 tickets are drawn without replacement. Find the probability that both tickets show an even number.
7/9
1/9
2/9
5/9

Q 4: Three boys Richard, Nick and Pete are waiters in a hotel. The respective probabilities of their breaking a dish in the kitchen are 1/3, 1/2 and 4/5. If one of them is selected at random, find the probability that a dish is broken.
49/90
23/87
16/111
67/98

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