[20分]app.math 概率

I assume that you understand the table that either only one sample is from P and one sample from Q, OR one sample from P and one sample from R are defective.

For case 1.

Since there are 3 samples from P, let's use the symbol G to denote good sample and D to denote defective sample.

Since there is only one defective out of the three samples from P, then these 3 samples can be :
GGD, GDG or DGG (This is where 3C1 comes from, one defective out of three). The probability for a sample from P being good = (1-15%) while the probability for a sample from P being defective =15%
Therefore total probability fot one defect from a sample of 3 is:
3C1 x (1-0.15)^2 x (0.15),i.e. sum of
GGD (1-0.15)x(1-0.15)x(0.15)
GDG (1-0.15)x(0.15)x(1-0.15)
DGG (0.15)x(1-0.15)x(1-0.15)


Similar for Q it is 3C1 x (1-0.2)^2 x (0.2)

For R since there are 4 samples from there and all of them are good, the probability is 4C0 x (1-0.1)^4

The same logic applies to case (2).

Hope this helps.

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