Question

The editor of a textbook publishing company is trying to decide whether to publish a proposed business statistics textbook. Information on previous textbooks published indicate that 10 % are huge​ successes, 20 % are modest​ successes, 40 % break​ even, and 30 % are losers.​ However, before a publishing decision is​ made, the book will be reviewed. In the​ past, 96 % of the huge successes received favorable​ reviews, 60 % of the moderate successes received favorable​ reviews, 50 % of the​ break-even books received favorable​ reviews, and 30 % of the losers received favorable reviews. Complete parts​ (a) and​ (b).

The probability that if the proposed textbook receives a favorable​ review, the book will be a huge success is?

What proportion of textbooks receives favorable reviews?

Answer 1

Let 'H' denote huge, 'M' modest , 'B' breakeven and 'L' losers.

For favourable 'F' and unfavourable 'U'. Using a decision tree is helpful. The first node is about the type of success. The next node is about the reviews. These are past reviews but conditional on the type of success. So these are conditional review. Eg: P(F | H) = 96%. The probability for unfavourable given same condtional event will be

1 - P(U |H) = 4%

Success		Reviews
		F	96%
Huge	10%
		U	4%
		F	60%
Modest	20%
		U	40%
		F	50%
BE	40%
		U	50%
		F	30%
Losers	30%
		U	70%

Using the conditional probability formula we can find the intersecting probability where

therefore

Therefore following probabilities are found by multiplying the previous ones

Eg: P(HF) = P(H) * P(F |H) = 10% * 96%

Event	P
P(HF)	0.096
P(HU)	0.004
P(MF)	0.120
P(MU)	0.080
P(BF)	0.200
P(BU)	0.200
P(LF)	0.090
P(LU)	0.210
Total	1

a.

The probability that if the proposed textbook receives a favorable​ review, the book will be a huge success is?

Here we want to know that the book will be a huges success given it has favourable review

P( H |F) =

The denominator probabilities takes all possible combination of favourable with different type of successs

P(F) = P(HF) + P(MF) + P(BF) + P(LF)

= 0.506

P(H | F) = 0.096 / 0.506

Ans: P(H |F) = 0.1897

this can be done using bayes' theorem where when the formula is simplified we reach the above conclusion

b.

What proportion of textbooks receives favorable reviews?

Again using the same formula we have different combinations for favourable revies

P(F) = P(HF) + P(MF) + P(BF) + P(LF)

ANS: P(F)= 0.506