Question

use the binomial expansion of (p+q)^n to calculate a binomial distribution given n=6 and p=0.3

Answer 1

please help!

Answer 2

Coefficients of raise to power 6 expressions according to Pascal's Triangle are 1,6,15,20,15,6,1

(p+q)^6 = p^6 + 6p^5 *q + 15p^4*q^2 + 20p^3*q^3 + 15p^2*q^4 + 6p*q^5 + q^6

When p = 0.3

7.29*10^-4 + 1.458q*10^-2 + 1.215q^2 *10^-1 + 0.54q^3 + 1.35q^4 + 1.8q^5 + q^6

Answer 3

Is this in probability?

Answer 4

yes @ranga and thnx @akshay1234

Answer 5

Then p represents success and equals 0.3
q represents failure and equals 1 - 0.3 = 0.7

To calculate the binomial distribution for n = 6, p = 0.3 and q = 0.7 you need to calculate the probability of exactly 0 success in 6 trials, exactly 1 success in 6 trials, .... exactly 6 successes in 6 trials.

Answer 6

im kinda still confused here haha

Answer 7

omg i still have no idea >.< so can we go step by step? im sorry! im a pain lol

Answer 8

\[\Large P(r) = ~^nCr(p)^r(q)^{n-r}\]\[\Large P(r) = ~^6Cr(0.3)^r(0.7)^{6-r}\]For exactly 0 success in 6 trials, put r = 0 and calculate the probability P(0)
For exactly 1 success in 6 trials, put r = 1 and calculate the probability P(1)
...
For exactly 6 success in 6 trials, put r = 6 and calculate the probability P(6)

Answer 9

I will just show one example of calculating, say, P(3). All I need to do is put r = 3 in the above formula and calculate the RHS. You can do the rest for r = 0 to 6.

Answer 10

ok

Answer 11

\[\Large P(r) = ~^6C_r(0.3)^r(0.7)^{6-r}\]\[\Large P(3) = ~^6C_3(0.3)^3(0.7)^{6-3} = \frac{6!}{3! * 3!}(0.3)^3(0.7)^3 = \]\[\Large \frac{6*5*4}{3*2}*0.027*0.343 = 0.18522\]

Answer 12

Calculate P(0), P(1), .... P(6) and list them. That will represent the binomial distribution probabilities of getting exactly 0 success in 6 trails, exactly 1 success in 6 trials, ....., exactly 6 successes in 6 trials, respectively.

Answer 13

now you understand what to do

Answer 14

yes thank u!