1. ## binomial

50 (a) $b(3;8,0.6) = \binom{8}{3}(0.6)^{3}(0.4)^{5} - \binom{8}{2}(0.6)^{2}(0.4)^{6}$

(b) $b(5;8,0.6) = \binom{8}{5}(0.6)^{5}(0.4)^{3} - \binom{8}{4}(0.6)^{4}(0.4)^{4}$

(c) $P(3 \leq X \leq 5) = \binom{8}{5}(0.6)^{5}(0.4)^{3} - \binom{8}{2}(0.6)^{2}(0.4)^{6}$

(d) $P(1 \leq X) = \sum_{x=1}^{12} \binom{12}{x}(0.1)^{x}(0.9)^{12-x}$

59. (a) $B(4;10,0.3) = 0.850$
(b) $b(4;10,0.3) = B(4;10,0.3) - B(3;10,0.3) = 0.2$
(c) $b(6;10,0.7) = B(6;10,0.7) - B(5;10,0.7) = 0.2$
(d) $P(2 \leq X \leq 4) = B(4;10,0.3) - B(1;10,0.3) = 0.701$
(e) $P(2 \leq X) = 1- P(X \leq 1) = 1 - B(1;10, 0.3) = 0.851$
(f) $P(X \leq 1) = B(1;10, 0.7) = 0$

(g) $P(2

60. The long-run percentage of defectives is $5 \%$. Also $X \sim \text{Bin}(25,.05)$.
(a) $P(X \leq 2) = B(2;25, .05) = 0.873$
(b) $P(X \geq 5) = 1-P(X \leq 4) = 1-B(4;25,.05) = 1-0.993 = 0.007$
(c) $P(1 \leq X \leq 4) = B(4;25,.05) - B(0;25,.05)$

these look right?

2. Originally Posted by shilz222
50 (a) $b(3;8,0.6) = \binom{8}{3}(0.6)^{3}(0.4)^{5} - \binom{8}{2}(0.6)^{2}(0.4)^{6}$
By definition:

(a) $b(3;8,0.6) = \binom{8}{3}(0.6)^{3}(0.4)^{5}$

RonL

3. Originally Posted by shilz222;106361 (c) [tex
P(3 \leq X \leq 5) = \binom{8}{5}(0.6)^{5}(0.4)^{3} - \binom{8}{2}(0.6)^{2}(0.4)^{6} [/tex]
If $X\sim B(8,0.6)$:

$P(3 \leq X \leq 5) = \binom{8}{5}(0.6)^{5}(0.4)^{3} +\binom{8}{4}(0.6)^{4}(0.4)^{4}+\binom{8}{3}(0.6)^ {3}(0.4)^{5}$

RonL