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.

idk

step-by-step explanation: