binomial, geo distribution

[匿名]

2013-03-10 9:33 am

1, in a number of questions, 50% of them are with 1 mark each, 30% of them are with 2 marks each and others are with 5 marks each. Mr Chan chooses questions randomly to test Anna. it is known the probability of Anna answering the questions with 1 mark, 2 marks and 5 marks each correctly are 0.9, 0.65, 0.3 respectively.

it is given that 5 questions with 1 mark each, 3 questions with 2 marks each and 2 questions with 5 marks each are chosen. find the probability that Anna answers exactly 5 questions correctly, where 4 of them carry 1 mark each. Answer = 0.0443

2, a shot is successful if it hits the center of a target. David will shoot at the target repeatedly til the first successful shot is made in each archery practice. suppose the shots are independent and the probability of having a successful shot is 0.4.

if the outcomes of two archery practices of David are independent, find the probability that 3 shots are required to make a successful shot in one practice, and 5 shots are required for another practice. Answer = 0.0149

3, there are 2 production lines A and B in a factory manufacturing cameras, which constitute 70% and 30% of the total production of the factory respectively. it is known that the production line A and B manufacture cameras independently, and the probabilities of manufacturing a camera in poor quality are 0.02 and 0.05 respectively.

the cameras are selected at random and checked one by one,

a) find the probability that 5th camera checked is the first camera found in poor quality. answer = 0.0258

b) if all 3 cameras checked are not in poor quality, what is the probability that at most 2 more cameras are required to be checked to find the first camera in poor quality? Answer = 0.0572

Please explain how you count the probabilities, thanks.

回答 (1)

andrew

2013-03-10 11:23 am

✔ 最佳答案

1.
5Q1M : 5 questions with 1 mark
3Q2M : 3 questions with 2 marks
2Q5M : 2 questions with 5 marks

√ : answer correctly
x : answer incorrectly

P(answer 4 Qs correctly in 5Q1M)
= P(4√ 1x in 5Q1M)
= 5C4 x (0.9)⁴ x (1 - 0.9)
= (5!/4!1!) x (0.9)⁴ x 0.1
= 0.32805

P(answer 1 Q correctly in 3Q2M or 2Q5M)
= P(1√ 2x in 3Q2M) x P(2x in 2Q5M) + P(3x in 3Q2M) x (1√ 1x in 2Q5M)
= [3C1 x 0.65 x (1 - 0.65)²] x (1 - 0.3)² + (1 - 0.65)³ x [2C1x 0.3 x (1 - 0.3)]
= 0.13505625

The required probability
= P(answer 4 Qs correctly in 5Q1M) x P(answer 1 Q correctly in 3Q2M or 2Q5M)
= 0.32805 x 0.13505625
≈ 0.0443

=====
2.
H : Hit the target
N : NOT hit the target
P(H) = 0.4
P(N) = 1 - 0.4 = 0.6

The required probability
= P(NNH in a practice) x P(NNNNH in another practice)
= 2C1 x [0.6² x 0.04] x [0.6⁴ x 0.04]
= 2 x 0.6⁶ x 0.4²
≈ 0.0149

=====
3.
(a)
A : production line A
B : production line B
P : a camera in poor quality
N : a camera not in poor quality

P(P)
= P(A and P) + P(B and P)
= 70% x 0.02 + 30% x 0.05
= 0.029

P(N)
= 1 - P(P)
= 1 - 0.029
= 0.971

The required probability
= P(NNNNP)
= 0.971⁴ x 0.029
≈ 0.0258

(b)
The required probability
= P([at most 2 more cameras to find the first P] | [NNN])
= [P(NNNP) + P(NNNNP)] / P(NNN)
= [0.971³ x 0.029 + 0.971⁴ x 0.029] / 0.971³
= 0.971³ x 0.029 x [1 + 0.971] /0.971³
= 0.029 x 1.971
≈ 0.0572

參考: andrew

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