## 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. 2 years ago 2 years ago

1. JayDS

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

2. ParthKohli

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

3. JayDS

yep

4. ParthKohli

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

5. ParthKohli

Was I right there?

6. JayDS

nope.

7. TransendentialPI

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

Firstly find all the combinations possible...

9. eliassaab

6C4

10. JayDS

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

11. waterineyes

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

Wait I can be wrong also...

13. eliassaab

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

14. JayDS

15. waterineyes

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

16. waterineyes

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

17. waterineyes

*come

18. JayDS

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

19. waterineyes

$1- \frac{20}{35} \implies \frac{3}{7}$ Sir's method is direct can you understand that by looking at the solution of him ??

20. JayDS

how is the 20 obtained?

21. waterineyes

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

22. waterineyes

Sorry $$^6C_3$$ it is on left hand side...

23. waterineyes

Getting or Not@ @JayDS

24. waterineyes

20/35 can be written as 4/7

25. JayDS

kk, so u used permutations there?

26. waterineyes

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

27. waterineyes

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

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

29. JayDS

kk.

30. waterineyes

Are you getting all ??

31. JayDS

well I get your working out.

32. JayDS

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

33. waterineyes

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

Nice call @eliassaab !

35. waterineyes

*yellow in place of zero there...

36. JayDS

but isn't it 6C3?

37. JayDS

and where does the 3 come from?

38. waterineyes

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

yep.

40. waterineyes

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

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

yep I get what you say

43. waterineyes

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

44. JayDS

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

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..