SCSA Mathematics Methods Discrete random variables

$E(X) = 0 \times 0.1 + 1 \times 0.2 + 2 \times 0.25 + 3 \times 0.35 + 4 \times 0.1 = 2.15$

The cost of making four birthday cakes is $4 \times ＄20 = ＄80$
The expected number sold is $E(X) = 2.15$, so the expected income is
$2.15 \times ＄50 = ＄107.50$
Thus, the expected profit for the shop is $＄107.50 - ＄80 = ＄27.50$

$Y = 0$ when all the birthday cakes are sold
So, the number of birthday cakes requested for sale is 3 or 4
So $P(Y = 0) = P(X = 3) + P(X = 4) = 0.35 + 0.1 = 0.45$

Using CAS

$E(X) = 0.8$

Using CAS

$\sigma(X) = 0.6$

$Var(X) = 0.36$

The expected payout for each game is

$＄5 \times \frac{4}{5} = ＄4$

Hence ＄4 is the minimum that the fair needs to charge so that their expected profit is non-negative.

You would need to return the first ball to the bag (and re-mix the balls) prior to the second ball being drawn. This would ensure that the two trials are independent and have the same probability of yielding a red ball.

$P(\text{fruit bruised})=0.03$

Determine the probability of a box containing bruised
fruit:

$P(\text{at least one bruised out of 4})$

$=1-P(\text{none bruised out of 4})$

$=1-(0.97)^4$

$=1-0.88529$

$=0.11471$

The expected number (mean) of boxes out of 140

$E(X)=np$

$=140\times 0.11471$

$=16.0594$

Answer: 16 boxes can be expected to contain bruised
fruit.

Sample B
Mean $= p = \frac{5}{10} = 0.5$

Variance $= p(1-p) = 0.5 \times 0.5$
$= 0.25$

Sample B has a larger variance than sample A.
Because sample B has the larger variance, it has more variability about the mean compared to sample A.

There are two outcomes: hit or miss.
Hit $P(\text{hit}) = \frac{1}{4}$
Miss $P(\text{miss}) = \frac{3}{4}$

Using $H$ to represent the number of 'hits'.
The mean of a binomial distribution is $np$.
$np = 20 \times \frac{1}{4} = 5$
The variance of a binomial distribution is $np(1-p)$.
$np(1-p) = 5 \times \frac{3}{4}$
$= \frac{15}{4} = 3\frac{3}{4}$

One condition is that there are only two outcomes for each selection: 'delayed' (success) or 'not delayed' (failure), i.e. Bernoulli trials.
Another condition that makes this context suitable is that the probabilities of each outcome do not change in each trial, i.e. $p = \frac{1}{5}$ and $q = \frac{4}{5}$

$n=3, p=\frac{1}{5}, 1-p=\frac{4}{5}$
$P(\text{at least } 2)$
$= P(X=2 \text{ or } X=3))$
$= C_2^3 p^2 (1-p)^1 + C_3^3 p^3 (1-p)^0$
$= 3 \left(\frac{1}{5}\right)^2 \times \left(\frac{4}{5}\right) + \left(\frac{1}{5}\right)^3$
$= 3 \times \frac{1}{25} \times \frac{4}{5} + \frac{1}{5^3}$
$= \frac{12}{125} + \frac{1}{125}$
$= \frac{13}{125}$

$X \sim \text{Bi}\left(4, \frac{3}{5}\right)$

$\Pr(X \geq 3)$

$= \Pr(X = 3) + \Pr(X = 4)$

$= \binom{4}{3}\left(\frac{3}{5}\right)^3\left(\frac{2}{5}\right) + \left(\frac{3}{5}\right)^4$

$= \frac{297}{625}$

$\Pr(X = 2 | X \geq 1) = \frac{\Pr(X = 2) \cap \Pr(X \geq 1)}{\Pr(X \geq 1)}$

$= \frac{\Pr(X = 2)}{1 - \Pr(X = 0)}$

$= \frac{6 \times \left(\frac{3}{5}\right)^2 \times \left(\frac{2}{5}\right)^2}{\left(\frac{5}{5}\right)^4 - \left(\frac{2}{5}\right)^4}$

$= \frac{6^3}{5^4 - 2^4}$

$\begin{aligned} P(2\le X\le 4)&=P(X\le 4)-P(X\le 1)\\ &=0.95-0.2\\ &=0.75 \end{aligned}$

$\begin{aligned} P(X=1\mid X\le 3)&=\dfrac{P(X=1)}{P(X\le 3)}\\ &=\dfrac{0.2}{0.6}\\ &=\dfrac{1}{3} \end{aligned}$

$E(X) = \sum p_i x_i$
$= \frac{1}{5} \times 0 + \frac{4}{5} \times 1$
$= \frac{4}{5}$

$Var(X) = \sum p_i (x_i - \mu)^2$
$= \frac{1}{5} \times \left(\frac{-4}{5}\right)^2 + \frac{4}{5} \times \left(\frac{1}{5}\right)^2$
$= \frac{4}{25}$

Standard deviation $= \sqrt{Variance}$
$= \frac{2}{5}$