EXT: Problems with Discrete Random Variables and Probabilities

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. 

Worked Examples

Example 1

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}}$5C43C08C4​

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)​=1214​1314​​=1213​

example 2

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​

 

A few more

Example 3

Example 4