Question

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.

Answer 1

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

Answer 2

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

Answer 3

yep

Answer 4

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

Answer 5

Was I right there?

Answer 6

nope.

Answer 7

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.

Answer 8

Firstly find all the combinations possible...

Answer 9

6C4

Answer 10

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

Answer 11

It will be:

\[^7C_4\]

Now find that yellow is one of them:

\[^1C_1 \times ^6C_3\]

Now subtract it from 1..

Answer 12

Wait I can be wrong also...

Answer 13

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

\]

Answer 14

@eliassaab got the right answer.

Answer 15

Do this:

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

Answer 16

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

Answer 17

*come

Answer 18

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

Answer 19

\[1- \frac{20}{35} \implies \frac{3}{7}\]

Sir's method is direct can you understand that by looking at the solution of him ??

Answer 20

how is the 20 obtained?

Answer 21

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

Answer 22

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

Answer 23

Getting or Not@ @JayDS

Answer 24

20/35 can be written as 4/7

Answer 25

kk, so u used permutations there?

Answer 26

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

Answer 27

C stands for combinations..

What you doing yet is all based on combinations..

There is selection only and not arrangement of things here...

Answer 28

\[\large ^nC_r = \frac{n!}{(n-r)! \times r!}\]

This is the general formula for finding Combinations...

Answer 29

kk.

Answer 30

Are you getting all ??

Answer 31

well I get your working out.

Answer 32

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

Answer 33

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..

Answer 34

Nice call @eliassaab !

Answer 35

*yellow in place of zero there...

Answer 36

but isn't it 6C3?

Answer 37

and where does the 3 come from?

Answer 38

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..

Answer 39

yep.

Answer 40

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??

Answer 41

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.

Answer 42

yep I get what you say

Answer 43

Write it as :

\[P(Yellow) \quad or \quad P(Y)\]

\[P(Not yellow) = P(\bar{Y})\]

Answer 44

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 :(

Answer 45

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..