Some of the books on a certain shelf are in English, and the rest of

Question

Some of the books on a certain shelf are in English, and the rest of the books are in Spanish. If 2 books are to be chosen at random from the shelf and neither book is returned to the shelf, what is the probability that at least one of the 2 books chosen will be English?

(1) On the shelf, the ratio of the number of books in Spanish to the number of books in English is 3:1.
(2) There are fewer than 20 books on the shelf.

(1) On the shelf, the ratio of the number of books in Spanish to the number of books in English is 3:1.
(2) There are fewer than 20 books on the shelf.

Bunuel · Accepted Answer

We can solve this question with formulas, however we could also step back and evaluate logically. Even when combining the statements together, we could have the number of Spanish and English books as 3 and 1; 6 and 2; 9 and 3; or 12 and 4. In each of these cases, the probability of getting at least one English book when choosing 2 books will be different. Hence, the answer is E.

Shakira876 · Answer

Can someone pls explain why cant the answer be A

Dalade · Answer

The trick is realising that because the selection is without replacement, whilst the first probability would be consistent for all multiples of the ratio (i.e., 1/4), it would change for the second selection depending on what multiples of the ratio we are dealing with. See example below
 
 9:3 = (3/12) + (2/11) 
 12:4 = (4/16) + (3/15)

SergejK · Answer

Understanding that we are dealing with at least probability is key. At least 1 of 2 books is English is equal to 1- probability none of the books are English = 1- probability of both being Spanish. As no books are replaced, we need to find the total number of books as ratios are not enough, as the scaling of those numbers will lead to different probabilities.

S1: S/E=3/1 -> we could have the case that s=9 and E=3. Another possibility is S=12, E=4. From this, we know that this can't be enough as 1-(12/16x11/15) is not the same as 1-(9/12x8/11) -> not sufficient. 
S2: Here, we are only told that the total number of books is less than 20 -> not sufficient

Both together can't be sufficient as in S1 we already accounted for the total being less than 20 and it was not sufficient so that both statements together are also not sufficient.

callingTardis · Answer

Number of spanish books = s Number of english books = e probability = 1 - both spanish = 1 - (choosing 2 from s/choosing 2 from e+s) = 1 - (SC2/E+SC2) 1. if s = 3;e = 1 ---> p = 0.5 if s = 6; e = 2 ----> p = 13/28 2. many possible values for s and e, resulting in different probabilities 1+2. s = 3x, e = x ---> s+e< 30 ---> 4x < 20 ---> x<5. x can be 1,2,3,4. we have already checked s = 1 and 2 in statement 1 therefore, E

Usernamevisible · Answer

Question / Doubt

Can I say that in probability question ratio is never sufficient to find actual probability ?
 
 Response

No.
A ratio is often sufficient in probability questions because probability depends on relative counts, not actual counts.
Example:
Spanish : English = 3 : 1
Could be:

* 3 Spanish, 1 English
* 30 Spanish, 10 English
* 300 Spanish, 100 English
 Probability of selecting one English book:
* 1/4
* 10/40 = 1/4
* 100/400 = 1/4
 Same probability.

---

For your specific question:
We need:
Probability(at least one English)
= 1 − Probability(both Spanish)
If Spanish : English = 3 : 1, let:

* Spanish = 3k
* English = k
 Then:
 Probability(both Spanish)
 = (3k / 4k) × ((3k−1)/(4k−1))
 Notice that k does not cancel completely because of the "-1" terms.
 Examples:
 k = 1:
* 3 Spanish, 1 English
* both Spanish = (3/4)(2/3) = 1/2
 k = 2:
* 6 Spanish, 2 English
* both Spanish = (6/8)(5/7) = 15/28
 Different probabilities.
 Therefore Statement (1) is not sufficient.

---

GMAT shortcut:
When sampling without replacement, ask:
"Will the ratio alone determine every probability term?"
If the expression contains:

* n−1
* n−2
* remaining objects after a draw
 then actual numbers may matter.
 When sampling with replacement (or a single draw), ratios usually determine the probability completely.
 So the rule is not:
 "Ratio is never sufficient."
 A better rule is:
 "Ratios are often sufficient, but in without-replacement problems the actual total may matter because of the subtraction terms."

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