Eighteen people on a amateur sports team are choosing a caption, a vice-caption, and a social
coordinate from among their ranks. How many ways are there to do this? State of this is
permutation or comdination and finish the problem?
Solution
as there is involvement of arrangement this is a state of permutation
answer:
choosing 3 out 0f 18 and arranging them in order of ranks
= 18C3 * 3! [ or18P3] = 4896.
Kodo Millet PPT made by Ghanshyam bairwa college of Agriculture kumher bhara...
Eighty percent of married couples paid for their honeymoon themselve.pdf
1. Eighty percent of married couples paid for their honeymoon themselves. You randomly select
20 married couples and ask if they paid for their honeymoon themselves. Find the probability
that the number of couples who say they paid for their honeymoon themselves is: a). exactly
fifteen b). at least twelve c). less than twelve
Solution
Here p = 0.8, q = 0.2, n = 20 Binomial distribution (p + q)^20 You want value for a)
15 b) 12 and above c) 11 and below For (a) exactly 15, you need 20C15 x (0.8)^15 x (0.2)^5.
This is the required term of the binomial expansion. 20C15 = 20C5 = 15,504 (multiply out and
see) (0.8)^15 = 0.03518 (0.2)^5 = 0.00032 The term when multiplied out = 0.1745 Answer to
part (a) = 0.1745 For (b) you have to sum all the terms for 12, 13, 14, 15, 16, 17, 18, 19, 20 i.e.
find 20C12, 20C13, 20C14 etc. then (0.8)^12 x (0.2)^8, (0.8)^13 x (0.2)^7, etc. which takes a bit
of time & patience. Some of the later terms get very small and can be neglected I guess. Once
you have got that answer, you subtract it from 1 to get the answer to part (c).