26% adults favor the use of unmanned drones by police agencies. Twelve U. S. adults are randomly selected. Find the probability that the number of US adults who favor the use of unmanned drones by police agencies is

(a) exactly three,

(b) at least four,

(C) less than eight.

P(3) =

(Round to three decimal places as needed)

a. Exactly three = 0.00117

b. At least four = 0.00457

c. Less than eight = 0.222

Step-by-step explanation:

a. Since 26% U.S adults favour unmanned drone by police agencies, if 12 such adults are selected at random, then the probability that exactly 3 of them will favour the use of the unmanned drones will be:

(26/100)^3 × (74/100)^9

This is because there's a 26% chance that the first 3 adults picked out of the randomly selected twelve will be in support of the unmanned drone and there's a 74% (100%-26%) chance that the other 9 (12-3) adults that will be picked will not be in support.

Therefore (26/100)^3 × (74/100)^9

= 0.0176 × 0.0665

= 0.00117

b. In this case, the probability that at least four U.S adults out of the randomly selected 12 will support the idea of an unmanned drone:

(26/100)^4

Now each of the four adults have a 26/100 chance of supporting the idea of an unmanned drone by police agencies. It doesn't matter what the other 8 adults support.

So the probability that at least four out of the 12 will be in support:

= (26/100)^4

= 0.00457

c. The probability that less than eight out of the randomly selected 12 will be in support means that at least 5 out of the selected 12 adults will be against the idea of an unmanned drone by police agencies. It doesn't matter what the other 7 are in support of. The 5 that will not be in support will have a 74/100 chance of being picked.

The probability that at least 5 non-supporters are picked

= (74/100)^5

= 0.74^5

= 0.222

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,