Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JayDS

  • 2 years ago

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.

  • This Question is Closed
  1. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. ParthKohli
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep

  4. ParthKohli
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. ParthKohli
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Was I right there?

  6. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nope.

  7. TransendentialPI
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  8. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Firstly find all the combinations possible...

  9. eliassaab
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    6C4

  10. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    It will be: \[^7C_4\] Now find that yellow is one of them: \[^1C_1 \times ^6C_3\] Now subtract it from 1..

  12. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Wait I can be wrong also...

  13. eliassaab
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @eliassaab got the right answer.

  15. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  16. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  17. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    *come

  18. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  19. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 ??

  20. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how is the 20 obtained?

  21. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  22. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  23. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Getting or Not@ @JayDS

  24. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    20/35 can be written as 4/7

  25. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    kk, so u used permutations there?

  26. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  27. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    C stands for combinations.. What you doing yet is all based on combinations.. There is selection only and not arrangement of things here...

  28. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\large ^nC_r = \frac{n!}{(n-r)! \times r!}\] This is the general formula for finding Combinations...

  29. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    kk.

  30. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Are you getting all ??

  31. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well I get your working out.

  32. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  33. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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..

  34. TransendentialPI
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Nice call @eliassaab !

  35. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    *yellow in place of zero there...

  36. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but isn't it 6C3?

  37. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and where does the 3 come from?

  38. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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..

  39. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep.

  40. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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??

  41. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  42. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep I get what you say

  43. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  44. JayDS
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 :(

  45. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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..

  46. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.