Question: Assume that all 7-digit numbers that do not begin with 0 or 1 are valid phone numbers. If a number is generated at random using this rule, what is the probability that this valid phone number would begin with a 3 and have a last digit that is even?
- 1/20
- 1/16
- 1/10
- 1/8
- 1/6
”Assume that all 7-digit numbers that do not begin with 0 or 1 are valid phone numbers.”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. GMAT Quant section consists of 31 questions in total. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.
Solution and Explanation
Approach Solution 1:
For a 7-digit valid phone number, we have one restriction - 0,1 cannot be the first digit.
Now, there are 8 possibilities for the first digit and 10 possibilities for the other 6 digits of the phone number.
Total possible phone numbers are 8*10^6
For a valid number which has to begin with a 3 and have the last digit that is even,
there is 1 possibility for the first digit and 5 possibilities for the last digit.
The possibility of finding such a valid phone number is 1*10^5*5
Probabality is :
1*10^5*5 / 8*10^6
=5 / 8*10
= 1/16
Therefore, the probability of having a phone number with 3 as first digit & an even last digit is 1/16.
Correct Answer: B
Approach Solution 2:
There are 8 available digits for the first number of the phone number;
For each of the next 5 numbers, there are 10 available digits,
For the final (even) number, there are 5 available digits.
The first digit must be 3; the 2nd through 6th numbers can be any of the 10 digits,
There are 5 digits that make the final number even.
Thus, the probability is
=1/8 x 10/10 x 10/10 x 10/10 x 10/10 10/10 x 5/10
= 1/8 x 1^5 x ½
= 1/16 x 1
= 1/16.
Correct Answer: B
Approach Solution 3:
For the first number of the phone number, there are 8 available digits; for each of the next 5 numbers, there are 10 available digits; and for the last (even) number, there are 5 available digits.
The first digit must be 3, and the second through sixth digits can be any of the other 10 digits. The final number is even since it has 5 even digits.
The probability is therefore 1/8 x 10/10 x 10/10 x 10/10 x 5/10 = 1/8 x 15 x 12 = 1/16 x 1 = 1/16.
In this scenario, the number of red marbles equals 2x=2 since x=1 and y=6.
There are just 6 options for answers.
Correct Answer: B
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