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 Group Title

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

2. ParthKohli Group Title

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

3. JayDS Group Title

yep

4. ParthKohli Group Title

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

5. ParthKohli Group Title

Was I right there?

6. JayDS Group Title

nope.

7. TransendentialPI Group Title

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 Group Title

Firstly find all the combinations possible...

9. eliassaab Group Title

6C4

10. JayDS Group Title

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

11. waterineyes Group Title

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 Group Title

Wait I can be wrong also...

13. eliassaab Group Title

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

14. JayDS Group Title

15. waterineyes Group Title

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

16. waterineyes Group Title

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

17. waterineyes Group Title

*come

18. JayDS Group Title

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

19. waterineyes Group Title

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

20. JayDS Group Title

how is the 20 obtained?

21. waterineyes Group Title

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

22. waterineyes Group Title

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

23. waterineyes Group Title

Getting or Not@ @JayDS

24. waterineyes Group Title

20/35 can be written as 4/7

25. JayDS Group Title

kk, so u used permutations there?

26. waterineyes Group Title

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

27. waterineyes Group Title

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 Group Title

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

29. JayDS Group Title

kk.

30. waterineyes Group Title

Are you getting all ??

31. JayDS Group Title

well I get your working out.

32. JayDS Group Title

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

33. waterineyes Group Title

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 Group Title

Nice call @eliassaab !

35. waterineyes Group Title

*yellow in place of zero there...

36. JayDS Group Title

but isn't it 6C3?

37. JayDS Group Title

and where does the 3 come from?

38. waterineyes Group Title

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 Group Title

yep.

40. waterineyes Group Title

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 Group Title

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 Group Title

yep I get what you say

43. waterineyes Group Title

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

44. JayDS Group Title

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 Group Title

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..