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JayDS
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.
why isn't it \[1-\frac{7C1}{7C4}\]?
\(\text{P(Not yellow) = 1 } - \text{P(Yellow)}\)
\(\text{P(Yellow)} = \Large {1 \over \binom{7}{4} }\) Maybe.
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.
Firstly find all the combinations possible...
yellow 1st,2nd,3rd and 4th
It will be: \[^7C_4\] Now find that yellow is one of them: \[^1C_1 \times ^6C_3\] Now subtract it from 1..
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...
ur right, so which one? =='' I hate these confusing questions and topics.
\[1- \frac{20}{35} \implies \frac{3}{7}\] Sir's method is direct can you understand that by looking at the solution of him ??
\[\large ^6C_4 = \frac{6!}{(6-3)! \times 3!} = \frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2} = 5 \times 4 = \boxed {\color{blue}{20}}\]
Sorry \(^6C_3\) it is on left hand side...
Getting or Not@ @JayDS
20/35 can be written as 4/7
kk, so u used permutations there?
\[1 - \frac{4}{7} \implies \frac{7-4}{7} \implies \large \boxed{\boxed {\frac{3}{7}}}\]
C stands for combinations.. What you doing yet is all based on combinations.. There is selection only and not arrangement of things here...
\[\large ^nC_r = \frac{n!}{(n-r)! \times r!}\] This is the general formula for finding Combinations...
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?
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..
Nice call @eliassaab !
*yellow in place of zero there...
and where does the 3 come from?
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..
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??
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.
Write it as : \[P(Yellow) \quad or \quad P(Y)\] \[P(Not yellow) = P(\bar{Y})\]
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 :(
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..