Question

Two digits are selected at random from the digits 1 through 9 i) If the sum is odd, what is the probability that 2 is one of the digit selected


ii) If 2 is one of the digits selected, what is the probability that the sum is odd?​

Answer 1

Answer:

Originally Answered: Two digits are selected at random from digits 1 through 9, if 2 is one of the digits selected, what is the probability that the sum is odd? The probability of the sum being even is 4/9, since only 4 of the nine numbers is even.

Answer 2

(i.) Therefore the probability that 2 is one of the digits selected if the sum is odd is 5/16.

(ii.) Therefore the probability that the sum is odd if 2 is one of the digits selected is 1/8.

Given:

Two digits are to be selected at random from digits 1 to 9.

To Find:

(i.)  If the sum is odd, what is the probability that 2 is one of the digits selected?

(ii.) If 2 is one of the digits selected, what is the probability that the sum is odd?​

Solution:

The given question can be solved very easily as shown below.

As there is no information regarding repetition, let us assume that repetition is not allowed.

Given that,

Two digits are to be selected at random from digits 1 to 9.

Let the probability of an odd number be P(A).

Let the probability of getting 2 as a digit be P(B).

Let the probability of getting 2 as one of the digits and also getting an odd number be P( A∩B ).

The number of ways of getting '2' as one digit is given,

⇒ 2   → If we fix the first digit as 2 then the 2nd digit can be any number from 1 to 9 except 2 because repetition is not allowed → So number of ways 8 ways.

⇒   2 → Similarly 8 ways

So the total number of ways of getting 2 as one of the digits = 8 + 8 = 16 ways

Total number of ways of selecting 2 numbers = 9. 8  = 72 ways

⇒ Hence probability of getting 2 in one of the digits P(B) = 16/72 = 4/18 = 2/9

Now the number of ways of getting odd numbers is given by,

To get an odd number, the units place should be filled with 1, 3, 5, 7, or 9

So the units digit has 5 ways to get filled and the tens digit has the remaining 8 numbers to get it filled.

So the number of ways in which odd numbers are obtained = 8. 5 = 40 ways

Total number of ways of selecting 2 numbers = 9. 8  = 72 ways

⇒ The probability of getting an odd number = P(A) = 40/72 = 5/9

Now the number of ways of getting 2 as one of the digits and also getting an odd number is given by,

If 2 is fixed in the tens place and we have 5 ways to fill the units place to get an odd number.

⇒ The probability of getting 2 as one of the digits and also getting an odd number = P( A∩B ) = 5/72 ways

(i.) If the sum is odd, what is the probability that 2 is one of the digits selected:

⇒ P(A/B) = P( A∩B )/P(B) = (5/72)/(2/9) = ( 5 × 9 )/ ( 2 × 72 ) = 5/16

Therefore the probability that 2 is one of the digits selected if the sum is odd is 5/16.

(ii.) If 2 is one of the digits selected, what is the probability that the sum is odd:

⇒ P(B/A) =  P( A∩B )/P(A) = (5/72)/(5/9) = ( 5 × 9 )/ ( 5 × 72 ) = 1/8

Therefore the probability that the sum is odd if 2 is one of the digits selected is 1/8.

(i.) Therefore the probability that 2 is one of the digits selected if the sum is odd is 5/16.

(ii.) Therefore the probability that the sum is odd if 2 is one of the digits selected is 1/8.

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