A batch of 200 calculators contains 3 defective units. What the probability that a sample of three calculators will have/be

a) no defective calculators

b) at least one defective calculator

c) all defective calculators

1. 0.048

2. 0.952

3. 0.6465

Step-by-step explanation:

Requirement 1

the probability of no defective calculator is 0.048

Given,

Total calculator = 200

Number of defective = 3

Probability of defective,

p = 3/200

= 0.015

And the probability of non-defective,

q = 1 - p

=1-0.015

=0.985

The Probability distribution of defective follows the normal distribution.

P [X=0] = 〖200〗_ (c_0) 〖 (.015) 〗^0 〖 (.985) 〗^ (200-0)

P [X=0] = 0.048

Requirement 2

the probability of no defective is 0.048.

here, the probability of at least one defective means minimum 1 calculator can be defective. So the probability of at least one will be the probability of less than or equal 1.

P [X = less than or equal 1] = 1 - P [X=0]

= 1 - 0.048

= 0.952

So the probability of at least one defective is 0.952.

Requirement 3

c) the probability of all defective calculators is 0.6465.

P [X=3] = P[X=0] + P[X=1] + P[X=2] + P[X=3]

Here,

P [X=0] = 0.048

P [X=1] = 〖200〗_ (c_1) 〖 (.015) 〗^1 〖 (.985) 〗^ (200-1)

= 0.148228

P [X=2] = 〖200〗_ (c_2) 〖 (.015) 〗^2 〖 (.985) 〗^ (200-2)

= 0.2245997

P [X=3] = 〖200〗_ (c_3) 〖 (.015) 〗^3 〖 (.985) 〗^ (200-3)

= 0.2257398

So, P [X=3] = 0.048+0.148228+0.2245997+0.2257398

= 0.6465675

So, the probability of all defective calculators is 0.6465.