Question

answer letter AND alde at Qoul 1- Choose the correct answer (Write down the correct For Non-of-the-above choice, write down the correct answer.) You purchase a certain product. The manual states that the lifetime T of the product, efined as the amount of time (in years) the product works properly until it breaks down, satisfies the following rule: P(T t) e for t2 0. What is the probability (rounded to nearest thousandth) that this device breaks down during the third year? (A 0.549 (B) 0.181 1 If the continuous random va variance of 75. The probability that X lies between 40 and 50 is: (C) 0.67 (D)0.122(E none of the above riable X is uniformly distributed with a mean of 45 and a - (E) none of the above 2pt. 1/3(C) 1/2 If a random variable X has a PDF fx(x) (D) 2/3 A) 1/4 lara ndom variable X has a PDF fx(x) ={2(x-1) 2 x elsewhere 1 ,then ECX) is: 0 pt. (B) 3/5 e. For three events A, B, and C, we know that A and C are independent, B and C are independent, A and B are disjoint, P(AUC)-2/3, P(BUC)-3/4, P(AUBÚC)-11/ 12, then P(A),P(B), and P(C) are respectively given by: (A)(B) (C2 (D) ,t, (E) none of the above 2'3'2 4'3'4 3'2'2 3'2(E) none of the above

With explanation please.

Answer 1

$a.~Probability~that~this~device~breaks~down~during~third~year\\ =P(T\geq 3)=exp(-3/5)=0.5488\approx 0.549\\ Option~(A).\\ b.~Suppose~X\sim~Uniform[a,~b]~then\\ E(X)=\frac{1}{b-a}\int_a^bxdx=\frac{a+b}{2}=45~or,~a+b=90.....(1)\\ E(X^2)=\frac{1}{b-a}\int_a^bx^2dx=\frac{a^2+ab+b^2}{3}\\ Var(X)=E(X^2)-E^2(X)=\frac{(b-a)^2}{12}=75~\\ or,~b-a=30....(2)~since~b>a\\ (1)+(2)\Rightarrow 2b=120\Rightarrow b=60\\ then~a=90-60=30\\ P(40\leq X\leq 50)=\frac{50-40}{60-30}=\frac{1}{3}\\ Option~B.\\ c.~E(X)=\int_1^22x(x-1)dx=\left[\frac{2}{3}x^3-x^2 \right ]^2_1=\frac{5}{3}\\ Option~A.\\$

$e.~Since~A~and~B~are~disjoint~so,\\ ~P(A\cup B\cup C)=P(A\cup C)+P(B\cap C^c)=11/12\\ or,~P(B\cap C^c)=11/12-2/3=1/4~or,~P(B)P(C^c)=1/4...(1),\\ ~since~B~and~C~are~independent\\ Similarly~P(A\cup B\cup C)=P(B\cup C)+P(A\cap C^c)=11/12\\ ~or,~P(A)P(C^c)=1/6....(2)\\ (2)~divided~by~(1)\Rightarrow P(A)=\frac{2}{3}P(B).....(3)\\ P(A\cup C)=P(A)+P(C)-P(A)P(C)~since~A~and~C~are~independent\\ Again~P(A\cup C)=2/3\Rightarrow P(A)+P(C)-P(A)P(C)=2/3\\ or,~\frac{2}{3}P(B)+P(C)-\frac{2}{3}P(B)P(C)=\frac{2}{3}=2P(B)P(C^c)+3P(C)=2....(4)\\ Similarly~P(B\cup C)=P(B)+P(C)-P(B)P(C)=\frac{3}{4}\\ or,~P(B)P(C^c)+P(C)=\frac{3}{4}....(5)\\ (4)-2\times (5)\Rightarrow P(C)=2-\frac{3}{2}=\frac{1}{2}\\ Now~from~(4),~2P(B)(1-1/2)+3/2=2\\ ~or,~P(B)=\frac{1}{2}~(using~P(C^c)=1-P(C)=1/2)\\ From~(3),~P(A)=\frac{1}{3}~since~P(B)=\frac{1}{2}\\ Option~C:~\frac{1}{3},\frac{1}{2},\frac{1}{2}$