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Sagot :
We want the probability of "at least one is defective". Let's call this "D"
D -> "at least one is defective"
So we want P(D).
However, notice that the probability of "at least one is defective" plus the probability of "no on is defective" is 100%.
Let's call "no one is defective" as "N". So:
[tex]\begin{gathered} P(N)+P(D)=1 \\ P(D)=1-P(N) \end{gathered}[/tex]We do this because it is much easier to calculate P(N).
The probability of "N", "no one is defective", is the probability of picking one of the good ones four times in a row. We have 18 defective and 35 good ones, total of 53. So the probability of picking a good one first is:
[tex]\frac{35}{53}[/tex]Now, we have 18 defective and 34 good ones, total of 52, so the probability of picking a good one now is:
[tex]\frac{34}{52}[/tex]The third is:
[tex]\frac{33}{51}[/tex]And the fourth is:
[tex]\frac{32}{50}[/tex]The probability of all this happening is P(N), and we can calculate by:
[tex]P(N)=\frac{35}{53}\cdot\frac{34}{52}\cdot\frac{33}{51}\cdot\frac{32}{50}\approx0.1788[/tex]Now, we can calculate P(D):
[tex]P(D)=1-P(N)\approx1-0.1788\approx0.821[/tex]So, the probability os at least one of the calculator is defective is 0.821, alternative D.
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