Victor Moses was trying to open a 12 digit lock set by his friend. He

Question

Victor Moses was trying to open a 12 digit lock set by his friend. He knew the first 8 digits of the code. Also, he knew that the last 4 digits were 4, 5, 7, 8 but not the order in which they should be entered at last 4 places. He dials a number at random keeping all the digits he knew in mind. What is the probability that the lock gets open on the third attempt?

A. 1/24
B. 1/8
C. 1/9
D. 4/35
E. 4/24

A. 1/24
B. 1/8
C. 1/9
D. 4/35
E. 4/24

RahulRajIN1988 · Answer

(1) P(First time wrong digits) =  \frac{(4!-1)}{(4!)} = \frac{(23)}{(24)}  ->All except the correct one
(2) P(Second time wrong digits) =  \frac{(22)}{(23)}  ->Minus the combo he tried in the first attempt
(3) P(Third time right digits) =  \frac{(1)}{(22)}   ->Gets the correct combo out of remaining 22 possible cases

P(needed) = (1)(2)(3) =  \frac{(1)}{(24)}

AndrewN · Answer

Fun Fact  :  Victor Moses  is the name of a former Chelsea F.C. player from the  2012-13, 13-14, 17-18, and 18-19 campaigns . Depending on when this "friend" set the combination, it could have been none other than the Blues legend Frank Lampard, who donned the number 8 in those earlier two seasons.

Oh, and I agree with the method that has been posted earlier. Happy studies, everyone.

- Andrew

Shikhar22 · Answer

And victor moses had a stint at liverpool on loan from chelsea and went back to chelsea only to play right wing back (initially a winger) and win the Premier league under Antonio Conte’s tutelage! 
😁

Posted from my mobile device

IanStewart · Answer

There are two different legitimate ways you could interpret the question. You could take it to mean "if Victor tries every possible code one at a time, what is the probability the lock opens specifically on his third attempt?" Then since there are 4! = 24 possible codes, each time Victor tries a new code (including the 3rd time), there will be a 1/24 chance it opens the lock.

But I'd probably interpret the question differently, as it's worded, because the wording seems to presuppose a third attempt will actually happen. So I'd interpret it to mean "If Victor is attempting to open the lock for the 3rd time, what is the probability it opens on that third attempt?" Then he's already tried two codes, so there are only 22 left to try, and the answer will be 1/22.

Shikhar22 · Answer

Hi Ian. Does the question not set us up to calculate the probability from the point of no attempt to finally getting it right on the 3rd attempt? In that case 1/24 seems okay.

Posted from my mobile device

Boho · Answer

4, 5, 7 and 8 can have 4!=24 arrangements so each try will have 1/24 chances of success; However, if we know first two tries have been unsuccessful and same combination is not repeated. we will have 22 unique combinations left for third attempt; Probability in that case will be 1/22.

(A) matches our first understanding.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Victor Moses was trying to open a 12 digit lock set by his friend. He

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Victor Moses was trying to open a 12 digit lock set by his friend. He

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards