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

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

yep

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

Was I right there?

nope.

Firstly find all the combinations possible...

6C4

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

Wait I can be wrong also...

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

@eliassaab got the right answer.

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

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

*come

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

how is the 20 obtained?

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

20/35 can be written as 4/7

kk, so u used permutations there?

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

kk.

Are you getting all ??

well I get your working out.

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

Nice call @eliassaab !

*yellow in place of zero there...

but isn't it 6C3?

and where does the 3 come from?

yep.

yep I get what you say

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