In a box of 15 tablets, 4 of the tablets are defective. three tablets are selected at random. what is the probability that a store buys three tablets and receives: a) no defective tablets, b) one defective tablet, and c) at least one non-defective tablet.

a) 0.394

b) 0.430

c) 0.981

Step-by-step explanation:

Use binomial probability:

P = nCr pʳ (1−p)ⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

and p is the probability of success.

Here, n = 3 and p = 4/15.

r is the number of defective tablets.

a) If r = 0:

P = ₃C₀ (4/15)⁰ (1−4/15)³⁻⁰

P = 1 (1) (11/15)³

P = 0.394

b) If r = 1:

P = ₃C₁ (4/15)¹ (1−4/15)³⁻¹

P = 3 (4/15) (11/15)²

P = 0.430

c) If r ≠ 3:

P = 1 − ₃C₃ (4/15)³ (1−4/15)³⁻³

P = 1 − 1 (4/15)³ (1)

P = 0.981

there would be 365 different 4 digit numbers

step-by-step explanation:

i belive that it is 365 because i asked google

2

step-by-step explanation: