A computer program generates a single digit by a random process, accor

Question

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

Kudos for a correct solution.

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

Lucky2783 · Accepted Answer

K (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9+ 1/10) =1

Note that 1/10 is least value and is equal to 0.1 so we can write above series as

K (1+0.5+0.3+0.25+0.2+0.16+0.5+residual) = 1

K=1/3+

P (3 or 5 or 7) =K*13/24

Required probability = 13/24 × 1/3+ = 4+/24 > 1/6

Answer should be B

P.s. 3+ denotes a value little greater than 3 . 
Same for 4+.

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

First, think of all the single digits: 0, 1, 2, … 8, 9. Now, the probability of generating any digit is “directly proportional to the reciprocal of one more than that digit.” So figure out that reciprocal for each digit:

0 gives 1/(0 + 1) = 1/1 = 1
1 gives 1/(1 + 1) = 1/2
2 gives 1/(2 + 1) = 1/3
3 gives 1/4
4 gives 1/5
…
9 gives 1/10

Now, these reciprocals are not themselves the probabilities—rather, the probabilities are “directly proportional” to those reciprocals. The easiest way to “rescale” the reciprocals so that they do correspond to the desired probabilities is to figure out the sum of all the reciprocals, then divide each reciprocal by that sum. The new, rescaled reciprocals will then automatically add up to 1, as probabilities must for a “complete” set of events (that is, exactly one of the events must happen—in this case, exactly one digit is generated).

What is the sum of 1, 1/2, 1/3, 1/4, through to 1/9? This would be a difficult sum to compute exactly, but notice that the problem asks for an answer “between” two benchmarks. That word “between” is a strong hint that you’ll need to approximate—and now’s the perfect time. To add up a bunch of fractions approximately, consider using decimal approximations. In this case, if you go to two decimal places, you’ll be plenty accurate.

1 + 0.5 + 0.33 + 0.25 + 0.2 + 0.17 + 0.14 + 0.13 + 0.11 + 0.1 = approximately what?

A quick vertical line-up and summation yields 2.93. So you’ll want to divide any reciprocal by 2.93, in order to figure out the actual probability for each digit.

Finally, the problem asks for the probability of generating an odd prime: that is, 3, 5, or 7. The reciprocals corresponding to those digits are 1/4, 1/6, and 1/8. The final approximate computation is this:

(1/4 + 1/6 + 1/8)/2.93 = approximately what?

The fractions add up to 13/24, or just a little more than 1/2. Dividing by 2.93 also gives you a result “just a little more” than dividing by 3 (or multiplying by 1/3). So you can conclude this:

(1/4 + 1/6 + 1/8)/2.93 = (13/24)/2.93 = (a little more than 1/2)(a little more than 1/3) = a little more than 1/6.

Given how close each approximation is to its lower benchmark, you can be nearly 100% sure that the result will not reach the next higher benchmark (1/3), which is twice as large as 1/6. In fact, the actual computation, rounded to 2 decimal places, yields 0.18, which is in fact only slightly above 1/6 and nowhere near 1/3.

The correct answer is B.

ScottTargetTestPrep · Answer

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

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  = 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

Kinshook · Answer

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

The probability of generating digit D = k* 1/(D+1) = k/(D+1)

The probability of generating digit 0 = k* 1/(0+1) = k 
The probability of generating digit 1 = k* 1/(1+1) = k/2
The probability of generating digit 2 = k* 1/(2+1) = k/3
The probability of generating digit 3 = k* 1/(3+1) = k/4 
The probability of generating digit 4 = k* 1/(4+1) = k/5 
The probability of generating digit 5 = k* 1/(5+1) = k/6 
The probability of generating digit 6 = k* 1/(6+1) = k/7
The probability of generating digit 7 = k* 1/(7+1) = k/8
The probability of generating digit 8 = k* 1/(8+1) = k/9
The probability of generating digit 9 = k* 1/(9+1) = k/10

Since the computer program generates a single digit by a random process, combined probability = 1

k + k/2 + k/3 + k/4 + .... + k/10 = 1
k (1+.5 +.3 + .25 + .2 + .17 + .14 + .13 + .11 + .1) = 1
2.9k = 1; 
k = 1/2.9 = 10/29 = .35 approx

Odd prime digits = {3,5,7}
The probability of generating odd prime digits = k/4 + k/6 + k/8 = (.25+.17+.125)k = .545*.35 = .19 approx
1/6 = .17; 1/3 = .33

IMO B

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