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NARA [144]
2 years ago
9

Fifteen percent of the students at Central High take Spanish, 12% take French, and 4% take both. If a student is selected at ran

dom, what is the probability that the student takes Spanish or French?
Mathematics
1 answer:
mel-nik [20]2 years ago
7 0

This question is incomplete because it lacks the appropriate options

Complete Question

Fifteen percent of the students at Central High take Spanish, 12% take French, and 4% take both. If a student is selected at random, what is the

probability that the student takes Spanish or French?

A 31%

B. 27%

C. 23%

D. 19%

Answer:

C. 23%

Step-by-step explanation:

At Central High

From the question, we are told that:

Spanish = 15%

French = 12%

Spanish and French = 4%

French or Spanish = unknown

Therefore, the probability that a student at random takes Spanish or French

= P(Spanish) + P(French) - P(Spanish and French)

= 15% + 12% - 4%

= 27% - 4%

= 23%

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(a) Probability that the second one selected is defective given that the first one was defective = 0.00450

(b) Probability that both are defective = 0.0112461

(c) Probability that both are acceptable = 0.986

    2. When Three containers are selected

(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451

(c) Probability that all three are defective = 6.855 x 10^{-8} .  

Step-by-step explanation:

We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.

                  Total containers = 445

                   Defective ones   = 3

           Non - Defective ones = 442 { Acceptable ones}

  • Two containers are selected, at random, without replacement from the batch.

(a) Probability that the second one selected is defective given that the first one was defective is given by;

  <em>Since we had selected one defective so for selecting second the available </em>

<em>   containers are 444 and available defective ones are 2 because once </em>

<em>    chosen they are not replaced.</em>

Hence, Probability that the second one selected is defective given that the first one was defective = \frac{2}{444} = 0.00450

(b) Probability that both are defective = P(first being defective) +

                                                                     P(Second being defective)

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(c) Probability that both are acceptable = P(First acceptable) +  P(Second acceptable)

Since, total number of acceptable containers are 442 and total containers are 445.

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(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;

<em>Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .</em>

Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = \frac{1}{443} = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;

<em>Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.</em>

Hence, Required probability = \frac{2}{443} = 0.00451 .

(c) Probability that all three are defective = P(First being defective) +

                              P(Second being defective) +  P(Third being defective)

        = \frac{3}{445}* \frac{2}{444}  * \frac{1}{443} = 6.855 x 10^{-8} .                

               

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