Aussie Maths & Science Teachers: Save your time with SmarterEd
The probability distribution table for a discrete random variable \(X\) is shown.
\begin{array}{|c|c|}
\hline \quad \quad x \quad \quad & \quad P(X=x) \quad\\
\hline 1 & 0.4 \\
\hline 2 & 0.2 \\
\hline 3 & \\
\hline
\end{array}
What is the value of \(P(X=3)\) ?
\(B\)
\(\text{Probabilities sum to 1:}\)
\(0.4+0.2+P(X=3) = 1\ \ \Rightarrow\ \ P(X=3) = 0.4\)
\(\Rightarrow B\)
A discrete random variable \(X\) is defined using the probability distribution below, where \(k\) is a positive real number.
\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{0}\ \ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{1}\ \ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{2}\ \ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{3}\ \ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ \ \text{4}\ \ \ \ \\
\hline
\rule{0pt}{2.5ex} \text{Pr} \ (X = x) \rule[-1ex]{0pt}{0pt} & 2k \rule[-1ex]{0pt}{0pt} & 3k \rule[-1ex]{0pt}{0pt} & 5k \rule[-1ex]{0pt}{0pt} & 3k \rule[-1ex]{0pt}{0pt} & 2k \\
\hline
\end{array}
Find \(\operatorname{Pr}(X<4 \mid X>1)\) (3 marks)
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\(C\)
\(2k+3k+5k+3k+2k=1\ \Rightarrow\ k=\dfrac{1}{15}\)
| \(\operatorname{Pr}(X<4 \mid X>1)\) | \(=\dfrac{\operatorname{Pr}(X>1)\ \cap \ \operatorname{Pr}(X<4)}{\operatorname{Pr}(X>1)}\) |
| \(=\dfrac{\operatorname{Pr}(1<X<4)}{\operatorname{Pr}(X>1)}\) | |
| \(=\dfrac{\frac{5}{15}+\frac{3}{15}}{\frac{5}{15}+\frac{3}{15}+\frac{2}{15}}\) | |
| \(=\dfrac{\frac{8}{15}}{\frac{10}{15}}\) | |
| \(=\dfrac{4}{5}\) |
A discrete random variable has probability distribution as shown in the table where `n` is a finite positive integer.
\begin{array} {|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ r\ \ \ & \ \ \ r^{2}\ \ \ & \ \ \ r^{3}\ \ \ & \ \ \ ...\ \ \ & \ \ \ r^{k}\ \ \ &\ \ \ ...\ \ \ & \ \ \ r^{n}\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & r^{n} & r^{n-1} & r^{n-2} & ... & r^{n-k+1} & ... & r \\
\hline
\end{array}
Show that `E(X) = n( 2r-1)`. (3 marks)
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`text(See Worked Solution)`
| `E(X)` | `= ∑x · P(X = x)` |
| `= r · r^n + r^2 · r^(n + 1) + … + r^n · r` | |
| `= r^(n + 1) + r^(n + 1) + … + r^(n + 1)` | |
| `= nr^(n + 1) …\ (1)` |
`text(Sum of probabilities = 1)`
`underset (text(GP where)\ a = r,\ r = r) (underbrace {r + r^2 + r^3 + … + r^(n-1) + r^n}) = 1`
| `(r(1-r^n))/(1-r)` | `= 1` |
| `r-r^(n + 1)` | `= 1-r` |
| `r^(n + 1)` | `= 2r-1 …\ (2)` |
`text{Substitute (2) into (1):}`
`E(X) = n(2r-1)`
The discrete random variable `X` has the probability distribution shown in the table below.
| `X = x` | `4` | `5` | `6` | `7` | `8` | |
| `P(x)` | `0.3` | `a` | `0.1` | `0.15` | `0.2` |
Find the value of `a`, and hence calculate the the expected value and variance of `X`. (3 marks)
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`2.31`
`0.3 + a + 0.1 + 0.15 + 0.2 = 1`
`=> \ a = 0.25`
`E(X) = ∑ x P(x)`
| `qquad X = x` | `4` | `5` | `6` | `7` | `8` | |
| `qquad P(x) qquad` | `0.3` | `0.25` | `0.1` | `0.15` | `0.2` | |
| `qquad x xx P(x) qquad` | `1.2` | `1.25` | `0.6` | `1.05` | `1.6` |
| `E(X)` | `= 1.2 + 1.25 + 0.6 + 1.05 + 1.6` |
| `= 5.7` |
| `text(Var)(X)` | `= E(X^2) – [E(X)]^2` |
| `= (4^2 xx 0.3) + (5^2 xx 0.25) + (6^2 xx 0.1) + (7^2 xx 0.15) + (8^2 xx 0.2) – 5.7^2` | |
| `= 2.31` |
Evaluate `p` and `q` in the discrete probability distribution table below, given that `E(X) = 3`. (3 marks)
\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 1\ \ \ & \ \ \ 2\ \ \ & \ \ \ 3\ \ \ & \ \ \ 4\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p & q & 0.2 & 0.4 \\
\hline
\end{array}
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`text(See Worked Solutions)`
| `p + q + 0.2 + 0.4` | `= 1` |
| `p + q` | `= 0.4\ \ …\ (1)` |
`text(Given)\ \ E(X) = 3,`
| `p + 2q + 0.6 + 1.6` | `= 3` |
| `p + 2q` | `= 0.8\ \ …\ (2)` |
`text{Subtract: (2) – (1)}`
| `q` | `= 0.4` |
| `:.p` | `= 0` |
The probability distribution of a discrete random variable, `X`, is given by the table below.
\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ & \ \ \ 1\ \ \ & \ \ \ 2\ \ \ & \ \ \ 3\ \ \ &\ \ \ 4\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.6p^{2} & 0.1 & 1-p & 0.1 \\
\hline
\end{array}
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a. `text(S)text(ince probabilities must sum to 1:)`
| `0.2 + 0.6p^2 + 0.1 + 1-p + 0.1` | `= 1` |
| `0.6p^2-p + 0.4` | `= 0` |
| `6p^2-10p + 4` | `= 0` |
| `3p^2-5p + 2` | `= 0` |
| `(p-1) (3p-2)` | `= 0` |
`:. p = 1 or p = 2/3`
| b.i. | `E(X)` | `= sum x * P (X = x)` |
| `= 1 xx (3/5 xx 2^2/3^2) + 2 (1/10) + 3 (1-2/3) + 4 (1/10)` | ||
| `= 4/15 + 1/5 + 1 + 2/5` | ||
| `= 28/15` |
| ii. | `P(X >= 28/15)` | `=P(X = 2) + P(X = 3) + P(X = 4)` |
| `= 1/10 + 1/3 + 1/10` | ||
| `= 8/15` |
The discrete random variable `X` has the probability distribution
\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ \ -1\ \ \ \ & \ \ \ \ \ 0\ \ \ \ \ & \ \ \ \ \ 1\ \ \ \ \ & \ \ \ \ \ 2\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p^{2} & p^{2} & \dfrac{p}{4} & \dfrac{4p+1}{8} \\
\hline
\end{array}
Find the value of `p.` (3 marks)
`1/2`
| `text(Sum of probabilities)` | `= 1` |
| `p^2 + p^2 + p/4 + (4p + 1)/8` | `= 1` |
| `16p^2 + 2p + 4p + 1` | `= 8` |
| `16p^2 + 6p – 7` | `= 0` |
| `(2p – 1) (8p + 7)` | `= 0` |
`:. p = 1/2,\ \ \ (p>0)`
The discrete random variable `X` has probability distribution as given in the table. The mean of `X` is 5.
\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ & \ \ \ 2\ \ \ & \ \ \ 4\ \ \ & \ \ \ 6\ \ \ &\ \ \ 8\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & a & 0.2 & 0.2 & 0.3 & b \\
\hline
\end{array}
The values of `a` and `b` are
`A`
`text(Sum of probabilities) = 1`
`a + 0.2 + 0.2 + 0.3 + b = 1`
`text(S)text{ince}\ E(X) = 5,`
| `5` | `=(0 xx a) + (2 xx 0.2) + (4 xx 0.2) + (6 xx 0.3) + 8b` |
| `8b` | `=2` |
| `:. b` | `=0.25` |
`:. a = 0.05,\ \ b = 0.25`
`=> A`