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Probability question

2022 · 27 Jul · Shift 1 · Q35
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  5. /2022 · 27 Jul · Shift 1 · Q35

Probability question

2022 · 27 Jul · Shift 1 · Q35

JEE MainMathematicsProbabilityMCQ+4 / −1
Let SSS be the sample space of all five digit numbers. It ppp is the probability that a randomly selected number from SSS, is a multiple of 7 but not divisible by 5 , then 9p9 p9p is equal to :
  1. A
    1.0146
  2. B
    1.2085
  3. C
    1.0285
  4. D
    1.1521
View written solutionFree

Correct answer: C

  1. Total number of five-digit numbers

The sample space SSS consists of all five-digit numbers, i.e. from 100001000010000 to 999999999999999.

Hence,

∣S∣=99999−10000+1=90000.|S| = 99999 - 10000 + 1 = 90000.∣S∣=99999−10000+1=90000.
  1. Count five-digit multiples of 7

We need numbers in [10000,99999][10000,99999][10000,99999] divisible by 777.

  • Smallest multiple of 777 greater than or equal to 100001000010000:
⌈100007⌉=1429\left\lceil \frac{10000}{7} \right\rceil = 1429⌈710000​⌉=1429

so the first multiple is

1429⋅7=10003.1429\cdot 7 = 10003.1429⋅7=10003.
  • Largest multiple of 777 less than or equal to 999999999999999:
⌊999997⌋=14285\left\lfloor \frac{99999}{7} \right\rfloor = 14285⌊799999​⌋=14285

so the last multiple is

14285⋅7=99995.14285\cdot 7 = 99995.14285⋅7=99995.

Therefore, the number of five-digit multiples of 777 is

14285−1429+1=12857.14285-1429+1 = 12857.14285−1429+1=12857.
  1. Count five-digit numbers divisible by both 7 and 5

A number divisible by both 777 and 555 must be divisible by

lcm⁡(7,5)=35.\operatorname{lcm}(7,5)=35.lcm(7,5)=35.

So we count five-digit multiples of 353535.

  • Smallest multiple of 353535 at least 100001000010000:
⌈1000035⌉=286\left\lceil \frac{10000}{35} \right\rceil = 286⌈3510000​⌉=286

thus first such multiple is

286⋅35=10010.286\cdot 35 = 10010.286⋅35=10010.
  • Largest multiple of 353535 at most 999999999999999:
⌊9999935⌋=2857\left\lfloor \frac{99999}{35} \right\rfloor = 2857⌊3599999​⌋=2857

thus last such multiple is

2857⋅35=99995.2857\cdot 35 = 99995.2857⋅35=99995.

Hence, the number of five-digit multiples of 353535 is

2857−286+1=2572.2857-286+1 = 2572.2857−286+1=2572.
  1. Count numbers divisible by 7 but not by 5

Required count:

12857−2572=10285.12857 - 2572 = 10285.12857−2572=10285.
  1. Compute the probability ppp
p=1028590000.p = \frac{10285}{90000}.p=9000010285​.
  1. Compute 9p9p9p
9p=9⋅1028590000=1028510000=1.0285.9p = 9\cdot \frac{10285}{90000} = \frac{10285}{10000} = 1.0285.9p=9⋅9000010285​=1000010285​=1.0285.
  1. Match with the options
9p=1.02859p = 1.02859p=1.0285

which matches Option C.

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