| X | 2 | 3 | 5 | 7 | 12 |
|---|---|---|---|---|---|
| P(X) | 3k | k | k | 2k | k |
The range of a discrete random variable X is $\{1, 2, 3\}$ and the probabilities are given by $P(X=1)=3k^3$, $P(X=2)=2k^2$, and $P(X=3)=7-19k$. Then $P(X=3)=$
Sum of probabilities must be 1:
$$ 3k^3 + 2k^2 + 7 – 19k = 1 \implies 3k^3 + 2k^2 – 19k + 6 = 0 $$Testing values (since $0 \le k \le 1$), for $k=1/3$:
$$ 3(\frac{1}{27}) + 2(\frac{1}{9}) – 19(\frac{1}{3}) + 6 = \frac{1}{9} + \frac{2}{9} – \frac{57}{9} + \frac{54}{9} = 0 $$So $k=1/3$. Then $P(X=3) = 7 – 19(1/3) = \frac{21-19}{3} = \frac{2}{3}$.
Among every 8 units of a product, one is likely to be defective. If a consumer orders 5 units, the probability that at most one unit is defective is:
$n=5$, $p=1/8$, $q=7/8$. We need $P(X \le 1) = P(0) + P(1)$.
$$ P(0) = \binom{5}{0}(\frac{7}{8})^5 = (\frac{7}{8})^5 $$ $$ P(1) = \binom{5}{1}(\frac{1}{8})(\frac{7}{8})^4 = 5(\frac{1}{8})(\frac{7}{8})^4 $$Total $= (\frac{7}{8})^4 [ \frac{7}{8} + \frac{5}{8} ] = (\frac{7}{8})^4 [\frac{12}{8}] = \frac{3}{2}(\frac{7}{8})^4$.
Two persons A and B play a game by throwing two dice. If sum is even, A gets $\frac{1}{2}$ point. If sum is odd, A gets 1 point. The arithmetic mean of A’s points is:
$P(\text{Even}) = 1/2$, $P(\text{Odd}) = 1/2$.
Mean $E[X] = (\frac{1}{2} \times \frac{1}{2}) + (1 \times \frac{1}{2}) = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}$.
A typist prepares a page with 1 error per 10 pages. In an assignment of 40 pages, if the probability of at most 2 errors is $p$, then $e^4 p =$
Mean $\lambda = 40/10 = 4$.
$$ p = P(X \le 2) = e^{-4} [ \frac{4^0}{0!} + \frac{4^1}{1!} + \frac{4^2}{2!} ] $$ $$ p = e^{-4} [ 1 + 4 + 8 ] = 13e^{-4} $$Therefore, $e^4 p = e^4(13e^{-4}) = 13$.
If $X$ is a random variable with probability distribution $P(X=k) = \frac{(2k+3)c}{3^k}$ for $k=0,1,2,\dots$, then $P(X=3)=$
Sum of probabilities is 1. Using AGP summation, we found $c = 1/6$.
$$ P(3) = \frac{(2(3)+3)(1/6)}{3^3} = \frac{9/6}{27} = \frac{1.5}{27} = \frac{1}{18} $$If a Poisson variate X satisfies $P(X=3)=P(X=5)$, then $P(X=4)=$
$\frac{\lambda^3}{3!} = \frac{\lambda^5}{5!} \implies \lambda^2 = 20 \implies \lambda = \sqrt{20}$.
$$ P(4) = \frac{e^{-\lambda}\lambda^4}{4!} = \frac{e^{-\sqrt{20}}(400)}{24} = \frac{50}{3}e^{-\sqrt{20}} $$A shelf has 3 math and 2 physics books. A student picks a book 3 times with replacement. The mean number of math books is:
Binomial distribution with $n=3$, $p = 3/5$.
Mean $= np = 3 \times \frac{3}{5} = \frac{9}{5}$.
In a Poisson distribution, if $\frac{P(X=5)}{P(X=2)} = \frac{1}{7500}$ and $\frac{P(X=5)}{P(X=3)} = \frac{1}{500}$, then the mean is:
Using the first ratio: $\frac{\lambda^5/120}{\lambda^2/2} = \frac{\lambda^3}{60} = \frac{1}{7500}$.
$$ \lambda^3 = \frac{60}{7500} = \frac{1}{125} \implies \lambda = \frac{1}{5} $$The probability distribution of X is given below. Then the standard deviation of X is:
| X | 2 | 3 | 5 | 7 | 12 |
| P(X) | 3k | k | k | 2k | k |
Sum $= 8k = 1 \implies k=1/8$.
Mean $\mu = 5$. $E(X^2) = 36$.
Variance $= 36 – 25 = 11$. SD $= \sqrt{11}$.
If the mean and variance of a binomial distribution are $\frac{4}{3}$ and $\frac{10}{9}$ respectively, then $P(X \ge 6)=$
$np=4/3$, $npq=10/9 \implies q=5/6, p=1/6, n=8$.
$P(X \ge 6) = P(6)+P(7)+P(8) = \frac{1}{6^8}[\binom{8}{6}5^2 + \binom{8}{7}5 + 1] = \frac{741}{6^8}$.
If three dice are thrown, then the mean of the sum of the numbers appearing on them is:
Mean of 1 die $= 3.5$.
Mean of 3 dice sum $= 3 \times 3.5 = 10.5$.
If $X \sim B(7,p)$ and $P(X=3) = P(X=5)$, then $p=$
$\binom{7}{3} p^3 q^4 = \binom{7}{5} p^5 q^2 \implies 35 q^2 = 21 p^2 \implies 5(1-p)^2 = 3p^2$.
Solving quadratic: $2p^2 – 10p + 5 = 0$. Since $p \le 1$, $p = \frac{5-\sqrt{15}}{2}$.
