Sum of the digits of the number on the card drawn being 10

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Problem 7

Each of the two digit numbers than can be formed from the integers 1 to 9 no digit repeated, is written on a card. Then all the cards are thoroughly shuffled and stacked. If one card is drawn from the stack, find the probability that the sum of the digits in the number on it will be 10.
Ans :
1
9

Solution

Number of two digit numbers that can be formed using the digits 1 to 9, without repeating the digits

⇒ Number of permutations of nine different things (digits) taking two at a time,

= 9P2

nPr = (n) × (n − 1) × (n − 2) × r times

= 9 × 8

= 72

We can check this by writing down the possible numbers.

(12), (13), (14), (15), (16), (17), (18), (19)

(21), (23), (24), (25), (26), (27), (28), (29)

...

(91), (92), (93), (94), (95), (96), (97), (98)

Experiment :

Drawing a card from the 72 cards marked with numbers formed using the digits 1 to 9

Total Number of Possible Choices

= Number of ways in which a card can be chosen from the 72 cards

⇒ n = 72C1
=
72
1
= 72

Let

A : the event of the sum of the digits of the number on the card drawn being 10

For Event A

Number of Numbers whose sum of the digits would be 10

= 8

{(19), (28), (37), (46), (64), (73), (82), (91)}

Favorable
(Sum of Digits is 10) 		Unfavorable
(Others) 		Total
Available 8 64 72
To Choose 1 0 1
Choices 8C164C072C1

Number of Favorable Choices

= Number of ways in which a card with a number whose sum of the digits is 10 can be drawn from the total 8 favorable cards

⇒ mA = 8C1
=
8
1
= 8

Probability of choosing a card with a number whose sum of the digits is 10

⇒ 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
=
8
72
=
1
9

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mAc = n − mA
= 72 − 8
= 64

in favor

Odds in Favor of choosing a card with a number whose sum of the digits is 10

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices

= mA : mAc

= 8 : 64

= 1 : 8

against

Odds against choosing a card with a number whose sum of the digits is 10

⇒ Odds against Event A

= Number of Unfavorable Choices : Number of Favorable Choices

= mAc : mA

= 64 : 8

= 8 : 1

Author : The Edifier
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