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JayDS
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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.
 one year ago
 one year ago
JayDS Group Title
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.
 one year ago
 one year ago

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JayDS Group TitleBest ResponseYou've already chosen the best response.0
why isn't it \[1\frac{7C1}{7C4}\]?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\(\text{P(Not yellow) = 1 }  \text{P(Yellow)}\)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\(\text{P(Yellow)} = \Large {1 \over \binom{7}{4} }\) Maybe.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Was I right there?
 one year ago

TransendentialPI Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Firstly find all the combinations possible...
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
yellow 1st,2nd,3rd and 4th
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
It will be: \[^7C_4\] Now find that yellow is one of them: \[^1C_1 \times ^6C_3\] Now subtract it from 1..
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Wait I can be wrong also...
 one year ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.0
\[\frac { 6C4}{7C4}= \frac 3 7 \]
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
@eliassaab got the right answer.
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Do this: \[1  \frac{^6C_3}{^7C_4}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
It will also comes out to be 3/7...
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
ur right, so which one? =='' I hate these confusing questions and topics.
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
\[1 \frac{20}{35} \implies \frac{3}{7}\] Sir's method is direct can you understand that by looking at the solution of him ??
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
how is the 20 obtained?
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
\[\large ^6C_4 = \frac{6!}{(63)! \times 3!} = \frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2} = 5 \times 4 = \boxed {\color{blue}{20}}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Sorry \(^6C_3\) it is on left hand side...
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Getting or Not@ @JayDS
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
20/35 can be written as 4/7
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
kk, so u used permutations there?
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
\[1  \frac{4}{7} \implies \frac{74}{7} \implies \large \boxed{\boxed {\frac{3}{7}}}\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
C stands for combinations.. What you doing yet is all based on combinations.. There is selection only and not arrangement of things here...
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
\[\large ^nC_r = \frac{n!}{(nr)! \times r!}\] This is the general formula for finding Combinations...
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Are you getting all ??
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
well I get your working out.
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
6!/(6−3)!×3! so n=6 and r=3 but where are these from?
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
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..
 one year ago

TransendentialPI Group TitleBest ResponseYou've already chosen the best response.0
Nice call @eliassaab !
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
*yellow in place of zero there...
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
but isn't it 6C3?
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
and where does the 3 come from?
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
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..
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
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??
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
anyway I am getting more confused and I mostly understand how to do all steps like 1Pr(yellow chosen) = Pr(yellow not chosen) but I just have trouble writing and doing the correct steps to get to the correct answer.
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
yep I get what you say
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Write it as : \[P(Yellow) \quad or \quad P(Y)\] \[P(Not yellow) = P(\bar{Y})\]
 one year ago

JayDS Group TitleBest ResponseYou've already chosen the best response.0
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 :(
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
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..
 one year ago
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