A-Level AQA Maths Mechanics Quantities, Units & Modelling: The discrete random variable X h

The discrete random variable X has the probability function $$ P(X = x) = \begin{cases} c(7 - 2x) & x = 0, 1, 2, 3 \\ k & x = 4 \\ 0 & \text{otherwise} \end{cases} $$ where c and k are constants - AQA - A-Level Maths Mechanics - Question 16 - 2021 - Paper 3

Question 16

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The discrete random variable X has the probability function $$ P(X = x) = \begin{cases} c(7 - 2x) & x = 0, 1, 2, 3 \\ k & x = 4 \\ 0 & \text{otherwise} \end{cases... show full transcript

Worked Solution & Example Answer:The discrete random variable X has the probability function $$ P(X = x) = \begin{cases} c(7 - 2x) & x = 0, 1, 2, 3 \\ k & x = 4 \\ 0 & \text{otherwise} \end{cases} $$ where c and k are constants - AQA - A-Level Maths Mechanics - Question 16 - 2021 - Paper 3

Step 1

Show that 16c + k = 1

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Answer

To show that $16c + k = 1$ , we first substitute the values of x into the probability function:

For $X = 0$ :

$P(X = 0) = c(7 - 2(0)) = 7c$
For $X = 1$ :

$P(X = 1) = c(7 - 2(1)) = 5c$
For $X = 2$ :

$P(X = 2) = c(7 - 2(2)) = 3c$
For $X = 3$ :

$P(X = 3) = c(7 - 2(3)) = c$
For $X = 4$ :

$P(X = 4) = k$

Since the total probability must equal 1, we can establish:

$7c + 5c + 3c + c + k = 1$

Simplifying gives:

$16c + k = 1$

This completes the proof.

Step 2

Given that P(X \geq 3) = 5/8 find the value of c and the value of k.

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Answer

Given that $P(X \geq 3) = \frac{5}{8}$ , we first express $P(X \geq 3)$ using the probability function:

$P(X \geq 3) = P(X = 3) + P(X = 4) = c + k$

We know from part (a) that:

$16c + k = 1$

Now, substituting into this equation:

$P(X \geq 3) = c + k = \frac{5}{8}$

From equation (1): $k = 1 - 16c$

Substituting this into equation (2):

$c + (1 - 16c) = \frac{5}{8}$

Combining terms gives: $1 - 15c = \frac{5}{8}$

Subtracting 1 from both sides: $-15c = \frac{5}{8} - 1 = \frac{5}{8} - \frac{8}{8} = -\frac{3}{8}$

Thus: $c = \frac{3}{120} = \frac{1}{40}$

Now substitute back to find k: $k = 1 - 16(\frac{1}{40}) = 1 - \frac{16}{40} = 1 - \frac{2}{5} = \frac{3}{5}$

Therefore:

$c = \frac{1}{40}$
$k = \frac{3}{5}$

The discrete random variable X has the probability function $$ P(X = x) = \begin{cases} c(7 - 2x) & x = 0, 1, 2, 3 \\ k & x = 4 \\ 0 & \text{otherwise} \end{cases} $$ where c and k are constants - AQA - A-Level Maths Mechanics - Question 16 - 2021 - Paper 3

Worked Solution & Example Answer:The discrete random variable X has the probability function $$ P(X = x) = \begin{cases} c(7 - 2x) & x = 0, 1, 2, 3 \\ k & x = 4 \\ 0 & \text{otherwise} \end{cases} $$ where c and k are constants - AQA - A-Level Maths Mechanics - Question 16 - 2021 - Paper 3

Show that 16c + k = 1

Given that P(X \geq 3) = 5/8 find the value of c and the value of k.

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