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egoroff_w [7]
4 years ago
7

Ex: Neckwear association reported 3% of ties sold in the US are bow ties. If 4 customers purchased a tie, find the probability t

hat at least one purchased a bow tie.
Mathematics
1 answer:
Elan Coil [88]4 years ago
8 0

Answer:

P=0.1147

Step-by-step explanation:

P(bow tie)=0.03

P(no bow tie)=0.97

n=4

P(at \ least \ one)=1-P(none)\\=1-(1-0.03)^4\\=0.1147

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Answer:

Required probability is 0.784 .

Step-by-step explanation:

We are given that in a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of the students passed both exams.

Let Probability that the students passed the first exam = P(F) = 0.74

     Probability that the students passed the second exam = P(S) = 0.72

     Probability that the students passed both exams = P(F \bigcap S) = 0.58

Now, if the student passed the first exam, probability that he passed the second exam is given by the conditional probability of P(S/F) ;

As we know that conditional probability, P(A/B) = \frac{P(A\bigcap B)}{P(B) }

Similarly, P(S/F) = \frac{P(S\bigcap F)}{P(F) } = \frac{P(F\bigcap S)}{P(F) }  {As P(F \bigcap S) is same as P(S \bigcap F) }

                          = \frac{0.58}{0.74} = 0.784

Therefore, probability that he passed the second exam is 0.784 .

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Answer:

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Step-by-step explanation:

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