Probability –what Formula to Use and a Brief Description of How to Do It?

A study is conducted to assess the impact of caffeine consumption, smoking, alcohol consumption, and physical activity on cardiovascular disease. Suppose that 40% of participants consume caffeine and smoke. If 8 participants are evaluated, what is the probability that:

* Exactly half of them consume caffeine and smoke?
* More than 6 consume caffeine and smoke?
* Exactly 4 do not consume caffeine or smoke?

Suggestion:

this is a binomial distribution with n = 8, p = .4, q = 1- p = .6
and formula for probability of k successes in n trials being

Pk] = nCk * p^k * q^(n-k) = n!/k!*(n-k)! * p^k * q^(n-k)

it will be clearer when we solve the 1st q

qa
P[4] = 8c4 * .4^4 * .6^4
= 8!/4!4! *.4^4 * .6^4
= 8*7*6*5 /(4*3*2*1) *.4^4 *.6^4
= 70 * .4^4 * .6^4
= .2322
———–

qb
P[>6] = P[7] + P[8] computed in the above manner
= 8c7 * .4^7 *.6^1 + 8c8 * .4^8 * .6^0
= 8*.4^7 * .6 + .4^8
= .0085
———–

qc
if 4 don't consume caffeine or smoke, (8 – 4) = 4 do.
we have already computed
P[4] = .2322
——————

note:
——-
note that what is defined as a "success" / "failure"
in a binomial distribution
is just a matter of convenience