Question

1.Suppose that, starting at a certain time, batteries coming off an assembly line are examined one by one to see whether they are defective (let D = defective and N = not defective). The chance experiment terminates as soon as a non-defective battery is obtained.

(a) Circle the set that gives 5 possible outcomes for this chance experiment.

{N, ND, NDN, NDND, NDNDN} {N, ND, NND, NNND, NNNND} {N, DN, DDN, DDDN, DDDDN} {D, DN, DNN, DNNN, DNNNN}

(b) Circle the outcomes that are in the event E, that the number of batteries examined is an odd number?

E = {DN, DDDN, DDDDDN, ...}
E = {N, DDN, DDDDN, DDDDDDN, ...}E = {D, DD, DND, DNND, DNNND}
E = {N, ND, NDN, NDND, NDNDN}
E = {D, DN, DND, DNDN, DNDND, ...}

2. Consider the chance experiment in which the type of transmission - automatic (A) or manual (M) - is recorded for each of the next two cars purchased from a certain dealer. Circle the set of all possible outcomes (the sample space).

{A, M, AM}
{A, M, AA, AM, MA, MM} {AA, AM, MM}
{A, M}
{AA, AM, MA, MM}

(c) Circle the outcomes in B, the event that at least one car has an automatic transmission

{AA, AM, MA} {MM}
{AM, MA} {AM, MA, MM}

5. Suppose that 2% of cereal boxes contain a prize and the other 98% contain the

message, "Sorry, try again." Consider the random variable x, where x = number of boxes

purchased until a prize is found. (Round all answers to four decimal places.)

0.0196

0.9412

0.0588

0.0004

0.0192 0.9210

(a) What is the probability that at most three boxes must be purchased?P(at most three boxes) =

(b) What is the probability that exactly three boxes must be purchased?P(exactly three boxes) =

(c) What is the probability that more than three boxes must be purchased?P(more than three boxes) =

Answer 1

Solution : ( 5 )

( a )

Probability that at most three boxes must be purchased is

P( atmost three boxes) = 0.02 + ( 0.98 * 0.02 ) + ( 0.98 * 0.98 * 0.02 )

= 0.02 + 0.0196 + 0.019208

= 0.058808

= 0.0588

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Solution : ( 5 )

( b )

Probability that exactly three boxes must be purchased

P( exactly three boxes ) = ( 0.98 * 0.98 * 0.02)

= 0.019208

= 0.0192

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Solution : ( 5 )

( c )

Probability that more than three boxes must be purchased

P( more than three boxes ) = 1 - 0.0588

= 0.9412