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

    • one year ago
  2. ParthKohli Group Title
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    \(\text{P(Not yellow) = 1 } - \text{P(Yellow)}\)

    • one year ago
  3. JayDS Group Title
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    yep

    • one year ago
  4. ParthKohli Group Title
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    \(\text{P(Yellow)} = \Large {1 \over \binom{7}{4} }\) Maybe.

    • one year ago
  5. ParthKohli Group Title
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    Was I right there?

    • one year ago
  6. JayDS Group Title
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    nope.

    • one year ago
  7. TransendentialPI Group Title
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    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
  8. waterineyes Group Title
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    Firstly find all the combinations possible...

    • one year ago
  9. eliassaab Group Title
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    6C4

    • one year ago
  10. JayDS Group Title
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    yellow 1st,2nd,3rd and 4th

    • one year ago
  11. waterineyes Group Title
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    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
  12. waterineyes Group Title
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    Wait I can be wrong also...

    • one year ago
  13. eliassaab Group Title
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    \[\frac { 6C4}{7C4}= \frac 3 7 \]

    • one year ago
  14. JayDS Group Title
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    @eliassaab got the right answer.

    • one year ago
  15. waterineyes Group Title
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    Do this: \[1 - \frac{^6C_3}{^7C_4}\]

    • one year ago
  16. waterineyes Group Title
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    It will also comes out to be 3/7...

    • one year ago
  17. waterineyes Group Title
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    *come

    • one year ago
  18. JayDS Group Title
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    ur right, so which one? =='' I hate these confusing questions and topics.

    • one year ago
  19. waterineyes Group Title
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    \[1- \frac{20}{35} \implies \frac{3}{7}\] Sir's method is direct can you understand that by looking at the solution of him ??

    • one year ago
  20. JayDS Group Title
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    how is the 20 obtained?

    • one year ago
  21. waterineyes Group Title
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    \[\large ^6C_4 = \frac{6!}{(6-3)! \times 3!} = \frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2} = 5 \times 4 = \boxed {\color{blue}{20}}\]

    • one year ago
  22. waterineyes Group Title
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    Sorry \(^6C_3\) it is on left hand side...

    • one year ago
  23. waterineyes Group Title
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    Getting or Not@ @JayDS

    • one year ago
  24. waterineyes Group Title
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    20/35 can be written as 4/7

    • one year ago
  25. JayDS Group Title
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    kk, so u used permutations there?

    • one year ago
  26. waterineyes Group Title
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    \[1 - \frac{4}{7} \implies \frac{7-4}{7} \implies \large \boxed{\boxed {\frac{3}{7}}}\]

    • one year ago
  27. waterineyes Group Title
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    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
  28. waterineyes Group Title
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    \[\large ^nC_r = \frac{n!}{(n-r)! \times r!}\] This is the general formula for finding Combinations...

    • one year ago
  29. JayDS Group Title
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    kk.

    • one year ago
  30. waterineyes Group Title
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    Are you getting all ??

    • one year ago
  31. JayDS Group Title
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    well I get your working out.

    • one year ago
  32. JayDS Group Title
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    6!/(6−3)!×3! so n=6 and r=3 but where are these from?

    • one year ago
  33. waterineyes Group Title
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    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
  34. TransendentialPI Group Title
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    Nice call @eliassaab !

    • one year ago
  35. waterineyes Group Title
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    *yellow in place of zero there...

    • one year ago
  36. JayDS Group Title
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    but isn't it 6C3?

    • one year ago
  37. JayDS Group Title
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    and where does the 3 come from?

    • one year ago
  38. waterineyes Group Title
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    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
  39. JayDS Group Title
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    yep.

    • one year ago
  40. waterineyes Group Title
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    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
  41. JayDS Group Title
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    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.

    • one year ago
  42. JayDS Group Title
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    yep I get what you say

    • one year ago
  43. waterineyes Group Title
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    Write it as : \[P(Yellow) \quad or \quad P(Y)\] \[P(Not yellow) = P(\bar{Y})\]

    • one year ago
  44. JayDS Group Title
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    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
  45. waterineyes Group Title
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    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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