# Consider 16 gallons of genetically modified potatoes. Suppose that 75% of these potatoes have an increased resistance to pests, 50% were engineered to have a longer shelf life, and 30% have an increased resistance to pests and were engineered to have a longer shelf life. If one of these potatoes is randomly selected, compute the probability that this potatoe has an increased resistance to pests, but was not engineered to have a longer shelf life.

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Potatoes have been genetically modified to increase resistance to pests and have a longer shelf life.

Of the potatoes with genetic modifications there is a 75% probability that the resistance to pests has been increased and a 50% probability that the shelf life has been increased. The probability that a potato has an increased resistance to pests and was engineered to have a longer shelf life is 30%.

Of all the potatoes there is a 0.75*0.3 = 22.5% probability that it has both the qualities.

The probability that a potato picked at random has an increased resistance to pests, but was not engineered to have a longer shelf life is 0.75*(0.7) = 0.525

The required probability is 52.5%

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## Related Questions

aruv | Student

Suppose that 75% of these potatoes have an increased resistance to pests, 50% were engineered to have a longer shelf life, and 30% have an increased resistance to pests and were engineered to have a longer shelf life. If one of these potatoes is randomly selected, compute the probability that this potatoe has an increased resistance to pests, but was not engineered to have a longer shelf life.

A=potatoes have an increased resistance to pests

P(A)=75%=.75

B=potatoes have a longer self life

P(B) =50 %=.50

AB= Potatoes have an increased resistance to pests and were engineered to have a longer shelf life

P(AB)  =30% =.30

`A barB=` Potatoes have an increased resistance to pests and were  not engineered to have a longer shelf life

`P(A barB)= ?`

where  we have difined following terms

`AnnB=AB`

`Ann barB=A barB`

Thus by total probability theorem

`P(A)=P(AB)+P(A barB)`

`.75-.30=P(A barB)`

`P(A barB)=.45`

Ans.