A computer program generates a single digit by a random process, according to which the probability of generating any

M7MBA wrote: ↑

Sun Jun 14, 2020 1:36 pm

A computer program generates a single digit by a random process, according to which the probability of generating any digit is directly proportional to the reciprocal of one more than that digit. If all digits are possible to generate, then the probability of generating an odd prime digit is between

A. 0 and 1/6
B. 1/6 and 1/3
C. 1/3 and 1/2
D. 1/2 and 2/3
E. 2/3 and 5/6

[spoiler]OA=B[/spoiler]

Source: Manhattan GMAT

Let's first forget about the condition given in the question and treat this question as

If all digits are possible to generate, then the probability of generating an odd prime digit is between

We know that there are 0-9: ten digits. Out of these 10 digits, there are 3 odd prime digits (3, 5 and 7).

So, the probability of choosing three digits out of 10 digits = 3/10 = 0.3

Looking ar the options, we see that 0.3 lies between 1/6 and 1/3 ( 0.167 and 0.33), so option B seems to be the correct answer.

But is it the correct answer? We must analyze further since 0.3, a probable value is too close to extreme value 0.33 of option B. In the light of option C (0.33 and 0.5), whose extreme value is also 0.33, is a close contender.

Now let's understand the question.

The question means that the probabilities of

1. The digit 0 \( \propto \frac{1}{(0 + 1)}= 1\)
2. The digit 1 \( \propto \frac{1}{(1 + 1)}= \frac{1}{(2)}\)
.
4. The digit 3 \( \propto \frac{1}{(3 + 1)}= \frac{1}{(4)}\)
.
6. The digit 5 \( \propto \frac{1}{(5 + 1)}= \frac{1}{(6)}\)
.
8. The digit 7 \( \propto \frac{1}{(7 + 1)}= \frac{1}{(8)}\)
.
10. The digit 9 \( \propto \frac{1}{(9 + 1)}= \frac{1}{(10)}\)

Note that the above fractions are not the values of probabilities, but the factors to which the probabilities are proportional to.

If we take '5' as the mean value of digits 0 to 9, we can take ignore the impact of the addition of '1' to its reciprocal. But the impacts of the addition of '1' on the reciprocal of '3' is more than that of '7'. So, the combined probability of getting 3, 5, or 7 would be slightly less than 0.3, so the correct answer would be option B.

Correct answer: B

Hope this helps!

-Jay
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M7MBA wrote: ↑

Sun Jun 14, 2020 1:36 pm

A computer program generates a single digit by a random process, according to which the probability of generating any digit is directly proportional to the reciprocal of one more than that digit. If all digits are possible to generate, then the probability of generating an odd prime digit is between

A. 0 and 1/6
B. 1/6 and 1/3
C. 1/3 and 1/2
D. 1/2 and 2/3
E. 2/3 and 5/6

[spoiler]OA=B[/spoiler]

Solution:

Since the sum of the probabilities of generating each single digit is 1, we can create the equation:

k + (1/2)k + (1/3)k + … + (1/10)k = 1

where k, (1/2)k, (1/3)k, …, (1/10)k are the probabilities of generating 0, 1, 2, …, 9, respectively.

We can factor out k from the equation above:

k[1 + 1/2 + 1/3 + … + 1/10] = 1

Let S = 1 + 1/2 + 1/3 + … + 1/10, so we have:

k * S = 1

k = 1/S

Now, let’s find a lower estimate and an upper estimate for S (note: the italic terms are less than the corresponding terms in S, the bold terms are greater and the regular ones are equal):

1/4 + 1/4 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + 1/16 < S < 1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8

1 + 1/2 + 1/8 < S < 1 + 1 + 1 + 3/8

(8 + 4 + 1)/8 < S < (8 + 8 + 8 + 3)/8

13/8 < S < 27/8

Since k = 1/S, we have:

8/27 < k < 8/13

The probability of generating an odd prime digit (i.e., 3, 5, and 7) is k[1/4 + 1/6 + 1/8] = k(13/24). If k is 8/27, then this probability is 8/27 * 13/24 ≈ 1/2 *1/3 = 1/6. If k is 8/19, then this probability is
8/13 * 13/24 = 1/1 * 1/3 = 1/3. We see that the desired probability is between 1/6 and 1/3.

Answer: B