probability

[rehman @ nayak rani]

The ace ,2,3,4,5,6,7,8 of spades are placed face up in a row on the table .Then a pack of 8 cards containing the ace1,2,3,4,5,6,7,8 of diamonds are shuffled and placed in front of the player .as each successive diamond is turned over the corresponding spade is removed from the row what is the probability that all the spades can be removed without any break occurring in the row of spades?
1)1/315
2)1/245
3)1/225
4)1/175

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Hi rehman,

What is the source of this question? I ask because it's not written in proper GMAT 'style', so if you're actually studying for the GMAT, you might want to consider working with more realistic practice materials.

The 'key' to dealing with this prompt is to realize that a card at either 'end' of the row can be selected at any 'step' - and the row will remain unbroken.

For example, the first card could be EITHER the Ace or the 8. If the Ace is first, then the second card could be EITHER the 2 or the 8. If the 8 is first, then the second could be EITHER the Ace or the 7. Etc.

Thus, there are always 2 ways that 'fit' what we want to have happen (except at the end when there's only 1 card left). That overall calculation would be:

(2/8)(2/7)(2/6)(2/5)(2/4)(2/3)(2/2) = 1/315

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

[rehman @ nayak rani]

The question was asked in Patna ims exam

[rehman @ nayak rani]

Two circles of equal radius are drawn to pass through the centre of one another .what is the probability that a point of the enclosed area chosen at random belong to the area common to the two circle
1)5/11
2)1/5
3)6/11
4)nota

[Matt@VeritasPrep]

The ace ,2,3,4,5,6,7,8 of spades are placed face up in a row on the table .Then a pack of 8 cards containing the ace1,2,3,4,5,6,7,8 of diamonds are shuffled and placed in front of the player .as each successive diamond is turned over the corresponding spade is removed from the row what is the probability that all the spades can be removed without any break occurring in the row of spades?
1)1/315
2)1/245
3)1/225
4)1/175

Since the cards are in a row, you'd have to pick them in this fashion:

A 2 3 4 5 6 7 8

2 3 4 5 6 7 8

3 4 5 6 7 8, etc.

OR

A 2 3 4 5 6 7 8

A 2 3 4 5 6 7

A 2 3 4 5 6, etc.


(Essentially you just always pick an end card, rightmost or leftmost.)


So you'd need to pick either the A or the 8 first, which has probability (1/4). From there, you'd need to pick either the rightmost or leftmost card, which has probability (2/7). Continue in this fashion, and you get

(1/4) * (2/7) * (1/3) * (2/5) * (1/2) * (2/3)

At this point there are only two cards left, and there can be no break in the row, so you're done.

[Matt@VeritasPrep]

Two circles of equal radius are drawn to pass through the centre of one another .what is the probability that a point of the enclosed area chosen at random belong to the area common to the two circle
1)5/11
2)1/5
3)6/11
4)nota

If I understand the setup correctly, the problem would be solved as follows.

Not sure what answer 4 means here, but that should be it; the ratio is not rational.