nomore asked in 科學及數學數學 · 1 decade ago

Maths probability 的問題

1.

Box A Box B

Total no. of bulbs 8 14

No. of defective bulbs 2 6

A box is chosen at random and then a bulb is drawn

If it is known that the bulb drawn is defective, find the probability that the box chosen is B.

Ans: 12/19

2. Calvin and Katherine take turns tossing a coin. The one who first gets a head will win the game. Suppose Katherine tosses the coin first.

Find the probability that Katherine wins the game.

Ans: 2/3

How to get the answer?

3. Probability of John, May, Jason wins the game are 0.8, 0.7, 0.6 respectively. Find the probablity that one of them wins the game.

Why the steps are like this?

0.8 + (1-0.8)(0.7) + (1-0.8)(1-0.7)(0.6)

Thanks so much!!

1 Answer

Rating
  • 1 decade ago
    Best Answer

    1.

    Denote:

    A: Box A is chosen

    B: Box B is chosen

    d: The bulb is defective

    P(A and d)

    = P(A) x P(d from Box A)

    = (1/2) x (2/8)

    = 1/8

    P(B and d)

    = P(B) x P(d from Box B)

    = (1/2) x (6/14)

    = 3/14

    P(d)

    = P(A and d) + P(B and d)

    = (1/8) + (3/14)

    = 19/56

    P(B | d)

    = P(B and d) / P(d)

    = (3/14) / (19/56)

    = (3/14) x (56/19)

    = 12/19

    =====

    2.

    Denote:

    H: A head is tossed

    T: A tail is tossed

    Katherine tosses in the 1st, 3rd, 5th ...... (2n - 1)th terms

    Katherine wins if:

    She tosses H in the 1st turn. Prob. = 1/2

    Or: T in the first 2 turns, but she tosses H in the 3rd turn = (1/2)2(1/2)

    Or: T in the first 4 turns, but she tosses H in the 5th turn = (1/2)4(1/2)

    .......

    P(Katherine wins)

    = (1/2) + (1/2)2(1/2) + (1/2)4(1/2) + (1/2)6(1/2) + .........

    = (1/2) + (1/4)(1/2) + (1/4)2(1/2) + (1/4)3(1/2) + ......

    It is the sum of a G.P. (or G.S)

    with first term a = 1/2 and common ratio r =

    Hence, P(Katherine wins)

    = (1/2) + (1/4)(1/2) + (1/4)2(1/2) + (1/4)3(1/2) + ......

    = a/(1 - r)

    = (1/2)/[1 - (1/4)]

    = (1/2)/(3/4)

    = (1/2) x (4/3)

    = 2/3

    =====

    3.

    Assume that John plays the game first. If John loses, May plays the game. If May also loses, Jason plays the game.

    P(John wins the game)

    = 0.8

    P(May wins the game)

    = P(John loses and then May wins)

    = P(John loses) x P(May wins)

    = (1 - 0.8) x 0.7

    P(Jason wins the game)

    = P(John loses, May loses, and Jason wins)

    = P(John loses) x P(May loses) x P(Jason wins)

    = (1 - 0.8) x (1 - 0.7) x 0.6

    P(one of them wins the game)

    = P(John wins the game) + P(May wins the game) + P(Jason wins the game)

    = 0.8 + (1 - 0.8) x 0.7 + (1 - 0.8) x (1 - 0.7) x 0.6

    = 0.8 + 0.14 + 0.036

    = 0.976

    =

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