A grocery store rewards card has a 7 digit number to identify the user. The firs

Question

A grocery store rewards card has a 7 digit number to identify the user. The first digit must be 1 or 2. The remaining six digits take values randomly between 0 – 9 inclusively.

a) How many different ID numbers are possible for this rewards card?

b) What is the probability that first three digits are all even (assume 0 is even)?

c) What is the probability that the first three digits contain at least 1 even number?

d) What is the probability that the ID number has no 3's or no 8's?

e) What is the probability that all of the first four digits are distinct (i.e. no digit repeated)? There are no restrictions on the remaining digits.

Explanation / Answer

a)

number of digits =N(for first digit 2 choice and for remaing 6 we have 10 choice for each)=2*106

=2000000

b)

probability that first three digits are all even =P(first is 2 for second and 3rd we have 5 choices adn for rest 3 10 choices each) =(1*53*103)/2000000=125/2000 =1/16=0.0625

c)probability that the first three digits contain at least 1 even number

=1-P(no even numbers) =1-P(first is 1 for second and 3rd we have 5 choices and for rest 3 10 choices each) =1-(1*53*103)/2000000= 15/16=0.9375

d)

probability that the ID number has no 3's or no 8's =P(for rest of 6 digit we have remaing 8 choices each) =(2*86)/2000000=0.262144

e)

probability that all of the first four digits are distinct =P(for first we have 2 choice for second remaing 9, for third 8 for fourth 7 and for remaing 3 10 choices)=(2*9*8*7*10*10*10)/2000000=0.504

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