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* Exact Half of them consume of Cafferine and smoke? x 0 1 2 3 4 5 6 7 8 n 8 8 8 8 8 8 8 8 8 *More than 6 consume caffeine and smoke? 23.224% THE

Please I need to understand the properly the calcuation which is the formula calcuation? thanks. Gloria

* Exact Half of them consume of Cafferine and smoke? x n p Probability of x successes = P(x) 0 8 0.4 0.01679616 1 8 0.4 0.08957952 2 8 0.4 0.20901888 3 8 0.4 0.27869184 4 8 0.4 0.023224 5 8 0.4 0.12386304 6 8 0.4 0.04128768 7 8 0.4 0.00786432 8 8 0.4 0.00065536 *More than 6 consume caffeine and smoke? 23.224% THE PROBABILITY THAT EXACTLY 4 OF THEM CONSUME CAFFEINE AND SMOKE x 0 0.16796 1 0.08958 2 0.209019 3 0.278692 4 0.23224 5 0.123863 6 0.041288 P(X<6)=p(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 0.991481 99.15% c. Exactly 4 do not consume caffeine or smoke? Ans. (1-0.23224)=0.76776 A. P( x successess=) (n! / x!(n-x)!) * ( P^x) * ( (1-P)^n-x) (applied to x=6 and x=3 in my calculation here) b. n=8 X=6 P 0.4 P(X)= 8! / 6! (8-6)! * (0.4^6) * (1-0.4)^(8- 6) P(X)=40320 / 1440 * (0.04096)*(.36) P(x)= 28 * .00147456 ANS = 0.23224 n=8 X=3 P 0.4 P(X)= 8! / 3!(8-3)! * (0.4^3) * (1-0.4)^(8- 3) P(X)= 40320/720 * (0.064)*(0.07776) ANS=0.27869184 4.129% THE PROBABILITY THAT EXACTLY 1 OF FEW don't CONSUME CAFFEINE AND SMOKE 27.87% the probablility is associated with three of them that consume both at most. x p convert to percentage 0 0.16796 x 100 = 16.796% 1 0.08958 x 100 = 8.958 % 2 0.209019 x 100 = 20.902 % 3 0.278692 x 100 = 27.87 % 4 0.23224 x 100 = 23.223 % 5 0.123863 x 100 = 12.39 % 6 0.041288 x 100 = 4.13 % Exactly 4 do not consume caffeine or smoke? x=3) (1- 0.278692) = 0.721308 0.721308 x100= 72.13% x=6) (1- 0.041288)= 0.958712 0.958712 x100=95.87% x=7(1-0.007786432) 99.22% x=8(1-0.00065536) 99.93% P(X)= 56 * .00497664
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