F.5 Maths Probability

2015-05-24 10:27 am

question: http://postimg.org/image/kdyoxeun5/

my calculation: http://postimg.org/image/5kk1cnmw1/

I want to ask part(c).
Let the 2 person be A and B.

The first method(ans=0.0417)is thinking the conditions:
(A,B):
≥2 , 0
≥1 , ≥1
0 , ≥2
But the answer is wrong.(correct ans:0.0432)

The second method(ans=0.9041)is thinking the conditions:
1-P(altogether = 0 or 1)
which (A,B):
0 , 0
1 , 0
0 , 1
But the answer is totally wrong@_@

I want to ask how to calculate the required probability by using the result of
(a) and (b)?
Thank you!!

回答 (1)

邊位都好

2015-05-24 5:32 pm

✔ 最佳答案

The prob. getting in part (a) and (b) are no use in this question. It is because :
The prob, that get no CD is :
P(0H,8D) + P(1H,7D) + P(2H,6D)
= C(8,0) * 0.8⁸ + C(8,1) * (0.2) * (0.8)⁷ + C(8,2) * (0.2)² * (0.8)⁶
= 0.7969The prob, that get 1 CD is :P(3H,5D) + P(4H,4D) + P(5H,3D)= C(8,3) * (0.2)³ * 0.8⁵ + C(8,4) * (0.2)⁴ * (0.8)⁴ + C(8,5) * (0.2)⁵ * (0.8)³= 0.2019The prob, that get 2 CDs is :
P(6H,2D) + P(7H,1D) + P(8H,0D)
= C(8,6) * (0.2)⁸ * 0.8² + C(8,7) * (0.2)⁷ * (0.8) + C(8,8) * (0.2)⁶= 0.001231
So, the prob. that they can altogether get at least 2 CDs is :P(A0,B2) + P(A1,B1) + P(A1,B2) + P(A2,B0) + P(A2,B1) + P(A2,B2)= 2*P(0CD,2CD) + 2*P(1CD,2CD) + P(1CD,1CD) + P(2CD,2CD)= 2*0.7969*0.001231 + 2*0.2019*0.001231 + 0.2019*0.2019 + 0.001231*0.001231= 0.0432 (corr to 4dp)

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