Question

In a psychological experiment, 20 children are each asked to solve a puzzle independently in a fixed amount of time. Interest lies in the number of children who succeed in completing the task. The experimental conditions show that the probability of completing the puzzle is 0.48.

a) Find the probability of exactly 6 children succeeding.

b) Find the probability of at least 2 children succeeding.

c) What is the average number of children succeeding?

d) After the puzzle, among 20 children we start to choose children one by one. What is the probability of choosing the first children who succeeded from the puzzle at the sixth choice?

Answer 1

Answer

This problem can be solved using a Binomial distribution.

We know that:

n= 20

p= 0.48

Also, for a Binomial distribution, we know that:
P(X=r)= nCr* p^r * q^(n-r)

a. P(X=6)

r=6

Thus, P(X=6)= 20C6* 0.48⁶ * 0.52⁽¹⁴⁾

= 0.05010464663

b. P(at least 2)

= 1- [P(X=0) +P(X=1)]

Thus,

= 1- [20C0* 0.48⁰ * 0.52⁽²⁰⁾ + 20C1* 0.48¹ * 0.52⁽¹⁹⁾ ]

= 1- [0.00000208962+0.00003857756]

= 1- [0.00004066718]

= 0.99995933282

c. average number of children succeeding

This is given by the formula n*p

= 20*0.48

= 9.6

d. P(first success at 6th try)

This would mean that the first 5 tries are failures and the 6th try is a success.

= 0.52⁵*0.48

= 0.01824979353