Math, asked by mayanksalunke1727, 3 months ago

A four-digit number is to be formed using the digits 0, 1, 2, 3, 4, 5. All the
digits are to be different. Find the probability that the digit formed is
(i) odd, (ii) greater than 4000, (iii) greater than 3400, and (iv) a multiple
of 5.

Answers

Answered by amitnrw
3

Given : A four-digit number is to be formed using the digits 0, 1, 2, 3, 4, 5

All the digits are to be different

To Find : the probability that the digit formed is

(i) odd, (ii) greater than 4000,

(iii) greater than 3400, and (iv) a multiple of 5

Solution:

digits 0, 1, 2, 3, 4, 5

case 1  :  1 of the digits is 0

rest 3 Digits can be selected in ⁵C₃ = 10 ways

Numbers can be be formed in  3 * 3!  = 18 ways  

= 18 * 10

= 180

case 2  : 0 is not the digit

rest 4 Digits can be selected in ⁵C₄ = 5 ways

Numbers can be be formed in  4!  = 24  ways  

= 24 * 5

= 120

Total Possible numbers  = 180 + 120  =  300

odd

Ending with 1 , 3 , 5 ( 3 ways)    Having 0   rest 2 number in ⁴C₂ = 6 ways

2 * 2! = 6 ways   hence 3 * 6 * 6 = 108

Ending with 1 , 3 , 5 ( 3 ways)    not having 0    rest 3 number in ⁴C₃ = 4 ways

3! = 6 ways   hence 3 * 4 * 6 = 72

= 108 + 72  = 180

Probability = 180/300  =  3/5

greater than 4000

1st Digit 4 or 5  2 ways Hence  2 * ⁵C₃ * 3! = 120 ways

120/300 = 2/5

greater than 3400

Having 1st digit 4 or 5  Hence  2 * ⁵C₃ * 3! = 120 ways

1st digit 3 , 2nd digit 4 or 5   1 * 2 *   ⁴C₂ * 2!  = 24 ways

Probability = 144/300  =  48/100    = 12/25

a multiple of 5

last Digit 0   then ⁵C₃ * 3!  = 60

last Digit 5  

then if 0 is one of the digits then ⁴C₂ * 2 * 2!  = 24  Ways

if 0 is not one of the digit then ⁴C₃ * 3!  = 24 ways

= 60 + 24 + 24 = 108

Probability  = 108/300  = 9/25

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