Sum of digits, Product of digits in the number on the card drawn

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Problem 3

One hundred tickets are numbered as 00, 01, 02, ... , 09, 10, 11, ... , 99. If one ticket is drawn at random from them and if A is the event of getting 9 as the sum of the numbers on the tickets and B the event of the product of the numbers on the card being 0, then P(A) = ? and P(B) = ?
Ans :
1
10
,
19
100

Solution

Total number of tickets

= 100

Experiment :

Drawing a ticket from the tickets marked from 00, 01 ..., 99

Total Number of Possible Choices

= Number of ways in which a ticket can be drawn from the total 100

⇒ n = 9C1
=
100
1
= 100

A : the event of the sum of the digits of the number on the ticket drawn being 9.

B : the event of the product of the digits of the number on the ticket drawn being 0.

For Event A

Number of tickets with numbers whose sum of the digits is 9

= 10 {(09), (18), (27), (36), (45), (54), (63), (72), (81), (90)}

Favorable Unfavorable
(Others) 		Total
Available 10 90 100
To Choose 1 0 1
Choices 10C190C0100C1

Number of Favorable Choices

= Number of ways in which a ticket with a number whose sum of the digits is 9 can be drawn from the total 10 favorable tickets

⇒ mA = 10C1
=
10
1
= 10

Probability of drawing a ticket with a number whose sum of the digits is 9

⇒ Probability of occurrence of Event A

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(A) =
mA
n
=
10
100
=
1
10

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mAc = n − mA
= 100 − 10
= 90

in favor

Odds in Favor of drawing a ticket with a number whose sum of the digits is 9

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices

= mA : mAc

= 10 : 90

= 1 : 9

against

Odds against drawing a ticket with a number whose sum of the digits is 9

⇒ Odds against Event A

= Number of Unfavorable Choices : Number of Favorable Choices

= mAc : mA

= 90 : 10

= 9 : 1

For Event B

Number of tickets with numbers whose product of the digits is 0

= 19

{(00), (01), (02), (03), (04), (05), (06), (07), (08), (09), (10), (20), (30), (40), (50), (60), (70), (80), (90)}

Favorable Unfavorable
(Others) 		Total
Available 19 81 100
To Choose 1 0 1
Choices 19C181C0100C1

Number of Favorable Choices

= Number of ways in which a ticket with a number whose product of the digits is 0 can be drawn from the total 19 favorable tickets

⇒ mB = 19C1
=
19
1
= 19

Probability of drawing a ticket with a number whose product of the digits is 0

⇒ Probability of occurrence of Event B

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(B) =
mB
n
=
19
100

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mBc = n − mB
= 100 − 19
= 81

in favor

Odds in Favor of drawing a ticket with a number whose product of the digits is 0

⇒ Odds in Favor of Event B

= Number of Favorable Choices : Number of Unfavorable Choices

= mB : mBc

= 19 : 81

against

Odds against drawing a ticket with a number whose prodct of the digits is 0

⇒ Odds against Event B

= Number of Unfavorable Choices : Number of Favorable Choices

= mBc : m

= 81 : 19

Author : The Edifier
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