What is the correct answers?

Question

1) A study of consumer smoking habits includes 186 people in the 18-22 age bracket (42 of whom smoke, 124 people in the 23-30 age bracket (40 of whom smoke),124 people in the 23-30 age bracket (40 of whom smoke,) and 99 people on the 31-40 age bracket (21 of whom smoke). if one person is randomly selected from this sample, find the probability of getting someone who is age 18-22 or does not smoke.

2) A bin contains 64 light bulbs of which 10 are defective. If 5 light bulbus are randomly selected from the bin with replacement, find the probability that all the bulbs selected are good ones. Round to the nearest thousandth if necessary.

3) In a batch of 8,000 clock radios 2% are defective. A sample of 11 clock radios is randonmly selected without replacement from the 8,000 and tested. The entire batch will be rejected if at least one of those tested is defective. What is the probability that the entire batch will be rejected?

4) The organizer of a television show must select 5 people to participate in the show. The participants will be selected from a list of 30 people who have written in the show. If the participants are selected randonmly, what is the probability that the 5 youngest people will be selected?

Answer 1

Step 1 :

4).

Total number of people is 30.

Five people are selected from the list of 30 pepple.

The participants are selected randomly.

For the event must choose an ordered sample of 5 from 30 people : = 30C₅.

Where 30 is the total number of possibilities to start and 5 is the number of selections made.

30C₅ = 30!/[(30 - 5)! * 5!]

        = 30!/[25! * 5!]

        = (30*29*28*27*26*25!)/[25! * 5!]

        = (30*29*28*27*26)/[5*4*3*2*1]

        = 29*7*27*26

        = 142506.

If there are exactly 5 youngest people and if they were all selected, the probability is 5/30C₅.

= 5/30C₅

= 5/142506

=  0.00003509.

The probability that the 5 youngest people will be selected is 0.00003509.

Solution :

The probability that the 5 youngest people will be selected is 0.00003509.

Answer 2

2)

Step 1:

The bin contains 64 light bulbs of which 10 are defective.

So there are 54 light bulbs in good condition.

If 5 light bulbs are randomly selected from the bin with replacement.

The probability of bulb in good condition is 54/64.

Let the five events be A , B , C , D  and E .

Probability of selecting first light bulb is P (A ) = 54/64

Probability of selecting second light bulb is P (B ) = 54/64

Probability of selecting third light bulb is P (C ) = 54/64

Probability of selecting fourth light bulb is P (D ) = 54/64

Probability of selecting fifth light bulb is P (E ) = 54/64

The five events are independent then P (A ∩B ∩C ∩D ∩E ) = P (A ). P (B ). P (C ).P (D ).P (E )

= (54/64)(54/64)(54/64)(54/64)(54/64)

= (54/64)⁵

= 0.427630

= 0.428.

Solution:

Probability of selecting 5 light bulbs selected are good ones is 0.428.

Answer 3

1)

Step 1:

Age group	Smokers	Non-smokers	Total
18-22	42	144	186
23-30	40	84	124
31-40	21	78	99

Probability of favorable events = (Number of favorable events)/(Total number of events)

Probability P (E ) = p (f  )/p (t )

To Find the probability of getting some one who is age 18-22 or does not smoke:

p (f  ) = 42 + 144 + 84 + 78

p (t ) = 186 + 124 + 99

The probability of getting some one who is age 18-22 or does not smoke is,

P (E ) = (42 + 144 + 84 + 78)/(186 + 124 + 99)

= 348/409

= 0.8508

Solution:

The probability of getting some one who is age 18-22 or does not smoke is 0.851.

Answer 4

Step 1 :

3).

Total number of clock radios is 8000.

2% are defective, means 160 are defective.

(8000 - 160) = 7840 are not defective.

Eleven clock radios are randomly selected from the list of 8000.

Probability(none are defective) = 7840C₁₁/8000C₁₁

7840C₁₁ = 7840!/[(7840 - 11)! * 11!]

             = 7840!/[7829! * 11!].

8000C₁₁ = 8000!/[(8000 - 11)! * 11!]

             = 8000!/[7989! * 11!].

7840C₁₁/8000C₁₁ = [ 7840!/[7829! * 11!] ] / [ 8000!/[7989! * 11!] ]

                         = [ 7840! * [7989! * 11!] ]/ [[7829! * 11!] * 8000! ]

                         = [ 7840! * 7989! ]/ [7829! * 8000! ]

                         = 0.8006.

Probability(none are defective) is 0.8006.

Probability(at least one defective) = 1 - Probability(none are defective)

                                                = 1 - 0.8006

                                                = 0.1994.

The probability that the entire batch will be rejected is 0.1994.

Solution :

The probability that the entire batch will be rejected is 0.1994.