Four colours are randomly picked from the 7 different colours of the rainbow. Calculate the probability that yellow will not be one of the colours chosen.

- anonymous

- schrodinger

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- anonymous

why isn't it \[1-\frac{7C1}{7C4}\]?

- ParthKohli

\(\text{P(Not yellow) = 1 } - \text{P(Yellow)}\)

- anonymous

yep

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## More answers

- ParthKohli

\(\text{P(Yellow)} = \Large {1 \over \binom{7}{4} }\)
Maybe.

- ParthKohli

Was I right there?

- anonymous

nope.

- anonymous

How about this
Number of ways that yellow could be the color:
7*6*5*4
Number of 4 color combos = 7C4
then divide the two and find the compliment.

- anonymous

Firstly find all the combinations possible...

- anonymous

6C4

- anonymous

yellow 1st,2nd,3rd and 4th

- anonymous

It will be:
\[^7C_4\]
Now find that yellow is one of them:
\[^1C_1 \times ^6C_3\]
Now subtract it from 1..

- anonymous

Wait I can be wrong also...

- anonymous

\[\frac { 6C4}{7C4}= \frac 3 7
\]

- anonymous

@eliassaab got the right answer.

- anonymous

Do this:
\[1 - \frac{^6C_3}{^7C_4}\]

- anonymous

It will also comes out to be 3/7...

- anonymous

*come

- anonymous

ur right, so which one? =='' I hate these confusing questions and topics.

- anonymous

\[1- \frac{20}{35} \implies \frac{3}{7}\]
Sir's method is direct can you understand that by looking at the solution of him ??

- anonymous

how is the 20 obtained?

- anonymous

\[\large ^6C_4 = \frac{6!}{(6-3)! \times 3!} = \frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2} = 5 \times 4 = \boxed {\color{blue}{20}}\]

- anonymous

Sorry \(^6C_3\) it is on left hand side...

- anonymous

Getting or Not@ @JayDS

- anonymous

20/35 can be written as 4/7

- anonymous

kk, so u used permutations there?

- anonymous

\[1 - \frac{4}{7} \implies \frac{7-4}{7} \implies \large \boxed{\boxed {\frac{3}{7}}}\]

- anonymous

C stands for combinations..
What you doing yet is all based on combinations..
There is selection only and not arrangement of things here...

- anonymous

\[\large ^nC_r = \frac{n!}{(n-r)! \times r!}\]
This is the general formula for finding Combinations...

- anonymous

kk.

- anonymous

Are you getting all ??

- anonymous

well I get your working out.

- anonymous

6!/(6−3)!×3! so n=6 and r=3 but where are these from?

- anonymous

See when we are asked to find the probability as in this case:
you are to find probability of not getting zero so you can do two things>
1. Find the probability of getting yellow and then subtract it from 1..
2. Other method is what @eliassaab gave you..
Out of 7 just delete yellow you are left with 6 colors..
so: \(^6C_4\) because you now choose the 4 colors out of 6 because you have deleted yellow color..

- anonymous

Nice call @eliassaab !

- anonymous

*yellow in place of zero there...

- anonymous

but isn't it 6C3?

- anonymous

and where does the 3 come from?

- anonymous

How you have to choose 4 out of 6 now..
You eliminated yellow but there will be still all the colors from 6 but not including yellow..
This is the method that Sir has given you..

- anonymous

yep.

- anonymous

I explain both the methods in more simpler way:
Suppose you have 2 colors Yellow and Green (My Favorite too)..
First One:
My method is too find yellow's probability and subtract it from 1..
So do that:
P(Yellow) = 1/2
P(Not Yellow) = 1 - 1/2 = 1/2
Second Method:
That Sir gave you:
Delete Yellow:
You are left with Green:
Find its probability: P(Not yellow) = P(Green) = 1/2
Both the answers will be same..
Getting??

- anonymous

anyway I am getting more confused and I mostly understand how to do all steps like 1-Pr(yellow chosen) = Pr(yellow not chosen) but I just have trouble writing and doing the correct steps to get to the correct answer.

- anonymous

yep I get what you say

- anonymous

Write it as :
\[P(Yellow) \quad or \quad P(Y)\]
\[P(Not yellow) = P(\bar{Y})\]

- anonymous

well it's late and I have to go to sleep, but thanks for the help and if I'll try go through it again myself, otherwise you can help me again tomorrow :(

- anonymous

See the steps I am giving to you:
Firstly find all the possible combinations : \(^7C_4 = 35\)
We know that:
\[P(Y) = 1 - P(\bar{Y})\]
So find P(Y) first:
\[P(Y) = ^6C_3 \times ^1C_1 = 20\]
So,
\[P(Y) = \frac{20}{35} = \frac{4}{7}\]
Just subtract it from 1 and you will get the required probability..

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