Question

A certain insecticide kills 80% of all insects in laboratory experiments. A sample of 14 insects is exposed to the insecticide in a particular experiment. What is the probability that more than 5 insects will survive? Round your answer to four decimal places. Answer How to enter your answer Tables Keypad Submit Answer

Answer 1

Solution:

Let X be a random variable which represents the number of insects survive in the laboratory experiment.

A certain insecticide kills 80% of all insects. It means 20% of the all insecticide survive.

Probability that an insect survive = 20/100 = 0.20

Let us consider survival of insects as success.

Hence, probability of success (p) = 0.2

Number of trials (n) = 14

Since, probability of success remains constant throughout the experiment, number of trials (n) is finite, there are only two outcomes of an trial (success and failure), therefore we can consider that X is binomial distributed random variable.

According to binomial probability law,

Probability of occurrence of exactly x successes in n trials is given by,

; x = 0, 1, 2, ..............., n

P(X = ) = (?)p"(1 – p)-1

We have to obtain P(X > 5). And we have n = 14 and p = 0.20

P(X > 5) = 1- P(X<5)

Using binomial probability law we get,

P(X > 5) = 1-PCX = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X >5) = 1- (0.2)'(1 – 0.2)14-0 + ( *)(0.2) (1 -0.2)14–1 + (0.2) (1 -0.2) 14-2 + (0.2)'(1 -0.2)14–3 + (0.2)^(1 – 0.2)14–4 + (19)(0.2)5(1 – 0.2)14–5

(0) -1. (0) - ਗ ) - 2) =1

[Since, n! = n(n-1)(n-2)(n-3)! , 2! = (2×1)=2, 3! = (3×2×1)=6, 4! = (4×3×2×1)=24 and 5! = (5×4×3×2×1)=120]

14! 14! 10 14! :. P(X > 5) = 1-(0.8)14 +14(0.2) (0.8)13 +; + (0.2) (0.8)12 + 14! (0 2310 8111 3!(14 - 3)(0.2)° (0.8)" + 02)4(0.8) (0.2) (0.8)" + m 2 (0.2)* + 51(14 – 5)! 0.8) -5)/(0.

14 x 13 x 12! :. P(X > 5) = 1-(0.8)14 + 14(0.2) (0.8)13 + + 14 x 13 x 12 x 11! (0.2) (0.8)" + - 14 x 13 x 12 x 11 x 10! (0.2)*(0.8)10 + (24) (10) (2)(12! 14 x 13 x 12 x 11 x 10 x 9! * (0.2)5(0.8) (120)(9!) (0:2)°(0.8)" + (6 (11!

:P(X > 5) = 1-(0.8) 4 + 14(0.2)(0.8)13 +91(0.2)-(0.8)2 + 364(0.2) (0.8) 11 + 1001(0.2)+(0.8)10 + 2002(0.2)(0.8)

:P(X > 5) = 1-(0.04398 +0.15393 +0.25014 +0.25014+ 0.17197 +0.08599)

::P(X > 5) = 1-0.9561 = 0.0439

Probability that more than 5 insects will survive is 0.0439.