Question

The landing of military fighter jets on aircraft carrier requires great skill, so on occasions it requires more than one attempt to achieve the landing. TOP GUN is a pilot who is assigned to an aircraft carrier and has a record of achieving 95 % of landings on aircraft carriers in the first attempt. In a particular exercise TOP GUN is assigned to make four (4) takeoffs and landings on the aircraft carrier to which it is assigned. Under the assumption that the resulting events in each landing attempt are statistically independent of each other determine:

Most import: without using the binomial distribution. *NO BINOMIAL DISTRIBUTION*

a) The probability that TOP GUN achieve four (4) landings in the first (1) try.

b) The probability that TOP GUN achieve at least one (1) landing out of the four (4) on the first try.

Answer 1

The probability that TOP GUN achieving of landing on aircraft carriers in the first attempts is 0.95

That is p = 0.95 and n = 4 takeoffs

a) P(TOP GUN achieve all 4 landings in the first try) that is P(X = 4)

P(X = 4) = P(achieving on first landing) * P(Second landing) * P(third landing) * P(fourth landing)

= 0.95 * 0.95 * 0.95 * 0.95 = 0.8145

The probability that TOP GUN achieve four (4) landings in the first (1) try is 0.8145

b) P(At least one landing of the four on the first try) that is

The at least probability can be written in compliment form as

P(X = 0) means TOP GUN did not achieve any landing out of the four on the first try.

The probability of landing is 0.95, therefore the probability of did not land is 1 - 0.95 = 0.05

P(X = 0) = P(first did not land) * P(second not ) * P(third not) * P(fourth did not land)

= 0.05 * 0.05 * 0.05 * 0.05 = 0.00000625

The probability that TOP GUN achieve at least one (1) landing out of the four (4) on the first try is 0.99999375