Four rods of lengths $1$, $2$, $3$ and $4$ units are placed in a bag from which one rod is selected at random. The probability of selecting a rod of length $l$ is $kl$. Find the value of $k$.
Show that the expected value of $X$, the length of the selected rod, is $3$ units and the variance of $X$.
After a rod has been selected it is not replaced. The probabilities of selection for each of the three rods remain are in the same ratio as they were before the first selection. A second rod is now selected from the bag. Defining $Y$to be the length of the rod and writing ${{P}_{1}}=\mathrm{P}\left( Y=1|X=2 \right)$, ${{P}_{2}}=\mathrm{P}\left( Y=2|X=1 \right)$ show that $16{{P}_{1}}=9{{P}_{2}}$
Show also that $\mathrm{P}\left( X+Y=3 \right)=\frac{17}{360}$.

Four rods of lengths $1$, $2$, $3$ and $4$ units are placed in a bag from which one rod is selected at random. The probability of selecting a rod of length $l$ is $kl$. Find the value of $k$.

Show that the expected value of $X$, the length of the selected rod, is $3$ units and the variance of $X$.

After a rod has been selected it is not replaced. The probabilities of selection for each of the three rods remain are in the same ratio as they were before the first selection. A second rod is now selected from the bag. Defining $Y$to be the length of the rod and writing ${{P}_{1}}=\mathrm{P}\left( Y=1|X=2 \right)$, ${{P}_{2}}=\mathrm{P}\left( Y=2|X=1 \right)$ show that $16{{P}_{1}}=9{{P}_{2}}$

Show also that $\mathrm{P}\left( X+Y=3 \right)=\frac{17}{360}$.

Discrete Random Variables (DRV) - General

Discrete Random Variables (DRV)

International Baccalaureate Higher Level Mathematics

IB HL Math

Discrete Random Variable - Probability Involving Proportionality

| $l$ | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| $P(L = l)$ | $k$ | $2k$ | $3k$ | $4k$ |

Since $\sum P(L = l) = 1$:
\[k + 2k + 3k + 4k = 1\]
\[10k = 1\]
\[k = \frac{1}{10}\]

\[E(X) = 1 \cdot k + 2(2k) + 3(3k) + 4(4k) = 30k = 30\left(\frac{1}{10}\right) = 3\]

\[E(X^2) = 1^2 \cdot k + 2^2(2k) + 3^2(3k) + 4^2(4k) = 100k = 100\left(\frac{1}{10}\right) = 10\]

\[\text{Var}(X) = E(X^2) - [E(X)]^2 = 10 - 3^2 = 1\]

$P_1 = P(Y = 1 \mid X = 2)$
\[= \frac{P(Y = 1 \cap X = 2)}{P(X = 2)} = \frac{\frac{1}{8} \times \frac{2}{10}}{\frac{2}{10}} = \frac{1}{8}\]

$P_2 = P(Y = 2 \mid X = 1)$
\[= \frac{P(Y = 2 \cap X = 1)}{P(X = 1)} = \frac{\frac{2}{9} \times \frac{1}{10}}{\frac{1}{10}} = \frac{2}{9}\]

\[\frac{P_1}{P_2} = \frac{\frac{1}{8}}{\frac{2}{9}} = \frac{9}{16} \implies 16P_1 = 9P_2\]

\[P(X + Y = 3) = P(X = 2 \cap Y = 1) + P(X = 1 \cap Y = 2)\]
\[= \frac{2}{10} \times \frac{1}{8} + \frac{1}{10} \times \frac{2}{9} = \frac{17}{360}\]

Discrete Random Variable - Probability Involving Proportionality

Discrete Random Variable - Probability Involving Proportionality

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