It is known that 10% of the calculators shipped from a particular factory are defective. In a random sample of ten calculators, the probability of no more than one defect is:

0.7361

Step-by-step explanation:

In this question we have

number to be 10

Then we have a probability of 10% = 0.10

We have q = 1-p

= 1-0.10 = 0.90

Then the probability of not more than 1 being defective:

P(x=0) + p(x= 1)

(10C0 x 0.1⁰ x 0.9^10-0)+(10C1 x 0.1¹ x 0.9^10-1)

= 1 x1 x0.3487 + 10 x 0.1 x 0.3874

= 0.3487 + 0.3874

= 0.7361

This is the the required probability and this answers the question.

probability = 10 percent = 0.1

q= 1- 10percent = 90% = 0.9

n = 4

To get the required probabiltiy for this question is

P(not greater than one is defective )=P(x=0)+P(x=1)

= 4C0x(0.1)⁰x(0.9)⁴+4C1x(0.1)¹x(0.9)³

= 0.9477

The required probability is 0.9477

i know the answer is 3

step-by-step explanation:

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5

2n=18

n=9