Assume that all 7-digit numbers that do not begin with 0 or 1 are vali

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?

A. 1/20
B. 1/16
C. 1/10
D. 1/8
E. 1/6

A. 1/20
B. 1/16
C. 1/10
D. 1/8
E. 1/6

pushpitkc · Answer

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 phone 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  \frac{1*10^5*5}{8*10^6} = \frac{5}{8*10} = \frac{1}{16}

Therefore, the probability of having a phone number with 3 as first digit & an even last digit is  \frac{1}{16}   (Option B)

Tulkin987 · Answer

7-digit number = ABCDEFG
Applying constraints, we can chose A out of 8 digits (exlude 0 and 1). 
So, Total Outcome =  8*10*10*10*10*10*10 = 8*10^6 
Desired Outcome =  1*10*10*10*10*10*5 = 5*10^5  
Because we can chose only one digit for A (that is 3) and 5 digits for G (even).

So,  P =  (5*10^5)/(8*10^6) = 1/16 .

Answer: B

JeffTargetTestPrep · Answer

The number of valid phone numbers that begin with a 3 and end with an even digit is:

1 x 10^5 x 5

The number of all valid phone numbers (i.e., the first digit is neither 0 nor 1) is:

8 x 10^6

So the probability is (10^5 x 5)/(8 x 10^6) = 5/80 = 1/16.

Alternate Solution:

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, and 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, and 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 1/2 = 1/16 x 1 = 1/16.
 
Answer: B

arshiya22 · Answer

So we have 1 out of 8 possibilities for number with 3 (excluding 0 and 1) and 8 possibilities for the last digit to be even

1/8*4/8= 1/16

Is that right?

mskx · Answer

Overall numbers after restriction: (9999-2000)+1 = 8000
numbers beginn with 3 = (3999-3000)+1 = 1000
even numbers within= 1000/2 = 500

500/8000 = 1/16  B

bumpbot · Answer

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Assume that all 7-digit numbers that do not begin with 0 or 1 are vali

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