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Westkost [7]
3 years ago
12

A new surgical procedure is said to be successful 60% of the time. Suppose the operation is performed five times and the results

are assumed to be independent of one another. What are the probabilities of these events? (Round your answers to three decimal places.) (a) all five operations are successful (b) exactly three are successful (c) less than two are successful
Mathematics
1 answer:
USPshnik [31]3 years ago
3 0

Answer:

0.07776, 0.3456,0.08704

Step-by-step explanation:

Given that a new surgical procedure is said to be successful 60% of the time

Let X be the no of successes

Then X has only two outcomes and each trial is independent of the other

Hence X is binomial with p = prob of success in a single trial = 0.60

n = Number of operations performed = 5

Prob that

a) all five operations are successful =

P(X=5)\\= (0.6)^5\\=0.07776

(b) exactly three are successful

=P(X=3) = 5C3 (0.6)^3(0.4)^2\\= 0.3456

(c) less than two are successful

==P(X

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The probability that a students likes pizza is,

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P(Student\ does\ not\ likes\ pizza)=\frac{No.\ of\ students\ who\ does\ not\ prefer\ pizza}{Total\ no.\ of\ students\ surveyed}

                                                   =\frac{19}{216} \\=0.088

The probability distribution of students who prefer different kinds of pizza is:

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       P(A\ Student\ prefers\ cheese)=\frac{No.\ of\ students\ who\ prefer\ cheese}{Total\ no.\ of\ students\ surveyed}

                                                       =\frac{43}{216}\\=0.199

  • The probability that a student likes sausage:

        P(A\ Student\ prefers\ sausage)=\frac{No.\ of\ students\ who\ prefer\ sausage}{Total\ no.\ of\ students\ surveyed}

                                                           =\frac{56}{216}\\=0.259

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       P(A\ Student\ prefers\ pepperoni)=\frac{No.\ of\ students\ who\ prefer\ pepperoni}{Total\ no.\ of\ students\ surveyed}  

                                                             =\frac{39}{216}\\=0.181

  • The probability that a student likes supreme:

       P(A\ Student\ prefers\ supreme)=\frac{No.\ of\ students\ who\ prefer\ supreme}{Total\ no.\ of\ students\ surveyed}

                                                           =\frac{28}{216}\\=0.130

  • The probability that a student likes another kind:

        P(A\ Student\ prefers\ another\ kind)=\frac{No.\ of\ students\ who\ prefer\ another\ kind}{Total\ no.\ of\ students\ surveyed}

                                                                   =\frac{31}{216}\\=0.144

Thus, the probability distribution table is displayed below:

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