Random Variables

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Problems with Discrete Random Variables and Probabilities

Lesson

Now that we're nearing the end of our work on general discrete random variables, let's take a look at two problems that collect together some of the key ideas about DRVs and probability.

A pencil case contains $5$5 blue pens and $3$3 red pens. $4$4 pens are drawn randomly from the pencil case without replacement.

Let $X$`X` represent the number of blue pens drawn.

(a) Describe the probability distribution for $X$`X`.

Think: Describing a probability distribution is the same as constructing a probability distribution. This particular situation will involve the use of combinations and counting techniques.

Do: We'll construct a table of values. Note we'll need at least 1 blue pen if we want a total of 4 pens.

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|

$P(X=x)$P(X=x) |
$\frac{\nCr{5}{1}\nCr{3}{3}}{\nCr{8}{4}}$5C13C38C4 |
$\frac{\nCr{5}{2}\nCr{3}{2}}{\nCr{8}{4}}$5C23C28C4 |
$\frac{\nCr{5}{3}\nCr{3}{1}}{\nCr{8}{4}}$5C33C18C4 |
$\frac{\nCr{5}{4}\nCr{3}{0}}{\nCr{8}{4}}$5 |

Now we'll simplify our table to give us these probabilities.

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|

$P(X=x)$P(X=x) |
$\frac{1}{14}$114 | $\frac{6}{14}$614 | $\frac{6}{14}$614 | $\frac{1}{14}$114 |

(b) Calculate the probability of between $1$1 and $4$4 blue pens being drawn from the pencil case.

Think: We can use our table to add all the relevant probabilities. Note that between $1$1 and $4$4 does not include $1$1nor $4$4.

Do: $P(X=2)+P(X=3)=\frac{6}{14}+\frac{6}{14}=\frac{12}{14}$`P`(`X`=2)+`P`(`X`=3)=614+614=1214

(c) $P(X>1|X<4)$`P`(`X`>1|`X`<4)

Think: In words, this question is asking what is the probability of more than $1$1 blue pen being drawn if we know less that less than $4$4 blue pens were drawn from the pencil case. We can use our result from part (b).

Do: $\frac{P\left(X=2\right)+P\left(X=3\right)}{P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)}=\frac{\frac{12}{14}}{\frac{13}{14}}=\frac{12}{13}$`P`(`X`=2)+`P`(`X`=3)`P`(`X`=1)+`P`(`X`=2)+`P`(`X`=3)=12141314=1213

When a student is completing a task set by their teacher on Spacemaths, the number of hints used is monitored by the system.

The probability of using at least $1$1 hint is $0.6$0.6

The probability of using $2$2 hints is the same as using $3$3 hints.

The probability of using $1$1 hint is the same as using $4$4 hints.

At most students can use $4$4 hints.

The probability they use $2$2 hints is half of the probability that they use no hints.

(a) Let $X$`X` represent the number of hints they used. Construct the probability distribution.

Think: Once again we will use a table to represent our distribution. We won't need combinations this time, but careful reading.

Do:

The first piece of information tells us that $P(X=0)=0.4$`P`(`X`=0)=0.4. From there we can deduce that $P(X=2)=0.2$`P`(`X`=2)=0.2. The rest comes together after that.

$x$x |
$0$0 | $1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|---|

$P(X=x)$P(X=x) |
$0.4$0.4 | $0.1$0.1 | $0.2$0.2 | $0.2$0.2 | $0.1$0.1 |

(b) What is the expected number of hints a student will use?

Think: Here we need to refer back to how to calculate the expected value.

Do: $E(X)=0\times0.4+1\times0.1+2\times0.2+3\times0.2+4\times0.1$`E`(`X`)=0×0.4+1×0.1+2×0.2+3×0.2+4×0.1

$E(X)=1.5$`E`(`X`)=1.5

(c) Given that a student used at least $2$2 hints, what is the probability they used $4$4 hints?

Think: We're dealing with conditional probability here and need to use these ideas to calculate our probability.

Do: $\frac{P\left(X=4\right)}{P\left(X\ge2\right)}=\frac{0.1}{0.2+0.2+0.1}=\frac{1}{5}$`P`(`X`=4)`P`(`X`≥2)=0.10.2+0.2+0.1=15

Two dice are rolled and the absolute value of the differences between the numbers appearing uppermost are recorded.

Complete the sample space.

Die $2$2 **1****2****3****4****5****6**Die $1$1 **1**$0$0 $\editable{}$ $\editable{}$ $3$3 $\editable{}$ $\editable{}$ **2**$1$1 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ **3**$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $2$2 $\editable{}$ **4**$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ **5**$4$4 $\editable{}$ $2$2 $\editable{}$ $\editable{}$ $\editable{}$ **6**$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Let $X$

`X`be defined as the absolute value of the difference between the two dice. Construct the probability distribution for $X$`X`using the table below.Enter the values of $x$

`x`from left to right in ascending order.$x$ `x`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $P$ `P`$($($X=x$`X`=`x`$)$)$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Calculate $P$

`P`$($($X<3$`X`<3$)$).Calculate $P$

`P`$($($X\le4$`X`≤4$|$|$X\ge2$`X`≥2$)$).

A probability distribution function is represented in the table below.

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|

$P$P$($($X=x$X=x$)$) |
$3k^2$3k2 |
$2k$2k |
$k$k |
$4k^2+k$4k2+k |
$2k$2k |

Write an equation to solve for the value of $k$

`k`.Calculate $P$

`P`$($($X\le4$`X`≤4$)$).Find $P$

`P`$($($X>1$`X`>1$|$|$X<4$`X`<4$)$).Find $E\left(X\right)$

`E`(`X`).

Investigate situations that involve elements of chance: A calculating probabilities of independent, combined, and conditional events B calculating and interpreting expected values and standard deviations of discrete random variables C applying distributions such as the Poisson, binomial, and normal

Apply probability distributions in solving problems