1) if a monkey is given 11 letters that can form the word

Question

1) if a monkey is given 11 letters that can form the word "probability", what is the probability that the monkey will form the word "probability"?

2)A player selects six numbers from 1 to 49. What is the probability of picking the six winning numbers?

3) A typewriter has 48 keys (including space, numbers, etc). What is the probability that a baby, who strikes keys randomly, will correctly type the word "hamlet"?

christoph · Answer

1) 1/11!
2) 6!/(49*48*47*46*45*44)
3) 1/48^6

will explain if correct...

christoph · Answer

1) 1/(11!/2!*2!)

2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.

mckenna · Answer

1) 1/(11!/2!*2!) = 4/11!
Correct.

2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
Correct.

Hey, this is same as your previous answer:
1/49C6 = 6! x 43! / 49! = 6! / (49x48x47x46x45x44)

Your previous answer was also correct.

christoph · Answer

oh right ! too late here in germany :-D

you can approach this problem with combination- as well as with permutation-reasoning...

rc1979 · Answer

Can you guys explain 1?

christoph · Answer

divide by 2! and 2! because there are 2 b`s and 2 `s. for example probabilty with the first b in 4th position and the second b in 6th position is 1 way and when you change the b`s you get a second way, but it is the same word. so you divide 2.

Antmavel · Answer

the number 2) was not so easy, but I got 4!/11! and 1/48^6

rc1979, christoph explained it well, there are 2 letters B (le'ts call it B1 and B2 for example) 
Usually you have to work with permutation concerning words problems because order normally matters, in that case there will be no difference if you choose B1 or B2, it will be the same word whereas each B is considered as a specific letter. So you'll have twice more chances with B and it's the same idea for I.

finally, normally for a word with 11 different litters the prob will be 1/11! but since you have 2 repetitions of 2 letters, it will be 1*2*2/11!

Hope it will help you to figure it out...

MA · Answer

it is a permutation not a combiation problem. so it is 
=1/(49P6)

1) if a monkey is given 11 letters that can form the word

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