Consider a student with an ID number: 13762348. The student is going to select two digits...

Consider a student with an ID number: 13762348. The student is going to select two digits at random from the digits in the ID number, one after another and without replacement. What is the probability that the sum of the two digits is less than 10, given that the first digit is an odd number?

Solutions

Expert Solution

We would consider both the 3's as different.

Given that a first digit is an odd number, the possible choices for an odd number are 1,3,7 or 3.

Given, that the first digit is an odd number, the probability that the first digit is 1 is = 1/4

If 1 is the first digit, from the rest of the 7 digits any number can be chosen and it would satisfy the condition that the sum of the two digits will be less than 10.

Hence, the probability that the sum is less than 10 given that the first digit is 1 is = 1

If 3 is the first digit, from the rest of the 7 digits 1,6,2,3 and 4 can be chosen and it would satisfy the condition that the sum of the two digits will be less than 10.

Hence, the probability that the sum is less than 10 given that the first digit is 3 is = 5/7

If 7 is the first digit, from the rest of the 7 digits 1 and 2 can be chosen in order to satisfy the condition that the sum of the two digits will be less than 10.

Hence, the probability that the sum is less than 10 given that the first digit is 7 is = 2/7

If 3 is the first digit, from the rest of the 7 digits 1,6,2,3 and 4 can be chosen and it would satisfy the condition that the sum of the two digits will be less than 10.

Hence, the probability that the sum is less than 10 given that the first digit is 3 is = 5/7

Hence, the total probability that the sum of the two digits is less than 10, given that the first digit is an odd number is

= 1/4 * (1 + 5/7 + 2/7 + 5/7)

= 1/4 * 19/7

= 19/28

= 0.679

Thank You!! Please Upvote!!

orchestra answered 1 year ago

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