Question

Four letters are randomly selected from the word ENCYCLOPAEDIA. Find the probability that one letter E will occur in the selection of 4 letters.

Answer 1

is answer 32/13

Answer 2

nope, it's probability so it has to be between 0 and 1

Answer 3

I don't know how to do this but let us find how many 4 letters words we can find from it ..


There are 13 letters but out of which E comes 2 times, C two times, A two times..

Is there any other letter repeating ??

Tell me..

Answer 4

you remember what we did yesterday, for selecting 6 cards from a deck with a queen for sure ??

can u apply same logic ??

Answer 5

you have set of 13 letters...

out of which you select 4 letters , including E for sure...

SO you have 2 E in the set...

getting it ??

Answer 6

@JayDS : dude, reply something... :D

Answer 7

uh yeh, let me think.

Answer 8

yep.

Answer 9

so, hows it gonna be ??

Answer 10

13C4?

Answer 11

yes, going good... that is main set..

Answer 12

that would be the denominator

Answer 13

perfect :)

Answer 14

Now lets work on Numerator..

it compromises of two events simultaneously..

Event of picking E from set * event of picking any three letters from the remaining letters (excluding E)

Answer 15

yep.

Answer 16

So it would look something like this..

Event of picking E from set * event of picking any three letters from the remaining letters (excluding E)

( 11C3 * 2C1 )

Answer 17

@Ganpat letters are repeating so we will use the special case here ..

Answer 18

special case, like ??

Answer 19

I don't get how you got 11C3 x 2C1 Ganpat.

Answer 20

See here some letters are repeating..

Suppose you have your name :

A is repeating 2 times..

So total words according to you is 6C4 if you have to make 4 letters word..

How??

I can write : Gapa

I can also exchange a here but you will get the same word.. Gapa although we have exchanged a here..

Answer 21

@waterineyes : I remember this.. but m not sure.. This should be used in case of Permutation ??

Answer 22

sorry:

We will do like this:

\[\frac{6!}{2!}\]

Answer 23

So, the probability for the evnt shall be,

( 11C3 * 2C1 ) / 13C4...

= 6/ 13...

as per ma calculations.. recheck...

Answer 24

You got the right answer @JayDS

Answer 25

why is it 11C3 x 2C1?

Answer 26

combination = order no matter
for one of the letters you are choosing an E from 2 E's so 2C1 and then u have to choose 3 letters from the remaining 11 letters.

Answer 27

and it's not 13C2?

Answer 28

you have 13 letters word.. and question specifically says E..
there are 2 E`s in the word..
So the way you can select E = 2C1

and now, you have to pick remaining 3 letters randomly from the remaining 11 letters.. So the event 11C3..

Answer 29

does that make sense ??

Answer 30

yep.

Answer 31

@waterineyes : This case u explained, i have a example for this.. (i guess is applicable..)

Arrange the letters in the word GANPAT.. such that no vowels come together..

as far as i reckon.. over here you have to use the example you showed.. whatsay ?

Answer 32

@Ganpat my concepts of P, C and Probability have gone vanished..

4 years before I had read all these things..

It is time to look them once more seriously..

I am stupid why I am using here P when it is just combinations..

Huh...

Answer 33

@waterineyes : Common man, no one is stupid... Knowledge, just gets updated over higher one.. glad u helped me to brush up my knowledge :) Thanks !!!

Answer 34

Just wondering, why does this not work?
I multiplied the probability of getting one "E" with the chances the drawing the next numbers.
2/13 x 11/12 x 10/11 x 9/10

Answer 35

You will be getting higher values than the real answer yes or no ??

Answer 36

Hmm. is the real answer 360? I got 0.115...

Answer 37

2/13 x 11/12 x 10/11 x 9/10..

The question says, one letter E will occur..
So,

2/13 - selecting a E from the set knowing there are two E..

So from remaining set, all the E should be excluded..
and you will be left with 11 letters, not 12...

Answer 38

hey, m sorry for this bad explanation.. actually doing parallel work.. so forgive..

Answer 39

Hm.. but then that makes the equation: 2/13 x 1x1x1

Answer 40

so thats not the right approach.. and can u explain from where does 1 * 1* 1 , comes from ??

Answer 41

since you said to exclude the remaining "E"s, the fraction I multiply after 2/13 would be 11/11 and after, 10/10, 9/9 which all make one. That's how I got the "1"s... maybe it doesn't make sense