Question

10. There are nine golf balls numbered from 1 to 9 in a bag. Three balls are randomly selected without replacement to form a 3-digit number.
a. How many 3-digit numbers can be formed? Explain your answer.
b. How many 3-digit numbers start with the digit 1? Explain how you got your answer.
c. What is the probability that the 3-digit number formed is less than 200? Explain your answer.

Answer 1

Answer:

a) 504

b) 56

c) 0.111

Step-by-step explanation:

Data provided in the question:

There are nine golf balls numbered from 1 to 9 in a bag

Three balls are randomly selected without replacement

a) 3-digit numbers that can be formed

= $^nP_r$

n = 9

r = 3

= ⁹P₃

= $\frac{9!}{9!-3!}$

= 9 × 8 × 7

= 504

b)  3-digit numbers start with the digit 1

=  _ _ _

in the above 3 blanks first digit is fixed i.e 1

we and we have 8 choices left for the last 2 digits

Thus,

n = 8

r = 2

Therefore,

= 1 × ⁸P₂

= 1 × $\frac{8!}{8!-2!}$

= 1 × 8 × 7

= 56

c) Probability that the 3-digit number formed is less than 200

Now,

The number of 3-digit number formed is less than 200 will be the 3-digit numbers start with the digit 1 i.e part b)

and total 3-digit numbers that can be formed is part a)

therefore,

Probability that the 3-digit number formed is less than 200

= 56 ÷ 504

= 0.111