IBDP Maths SL 4.7 Discrete and continuous random variables AA HL Paper 1- Exam Style Questions- New Syllabus

(a)

Since probabilities must be between 0 and 1, each probability must be non-negative.

\( 0.6-2a\ge0,\qquad 3a\ge0,\qquad 0.4-a\ge0 \)

These inequalities give

\( a\le0.3,\qquad a\ge0,\qquad a\le0.4. \)

Combining them,

\( \boxed{0\le a\le0.3} \)

Notice that the probabilities always sum to 1, so only the non-negativity conditions are required.

✅ Answer: \(\boxed{0\le a\le0.3}\)

(b)

When \(a=0.2\), the distribution becomes

\( P(X=1)=0.2,\qquad P(X=2)=0.6,\qquad P(X=3)=0.2. \)

Using the property

\( \mathrm{Var}(2-X)=\mathrm{Var}(X), \)

we first calculate the mean:

\( E(X)=1(0.2)+2(0.6)+3(0.2)=2. \)

Next, calculate

\( E(X^2)=1^2(0.2)+2^2(0.6)+3^2(0.2) =0.2+2.4+1.8=4.4. \)

Therefore,

\( \mathrm{Var}(X) =E(X^2)-[E(X)]^2 =4.4-2^2 =4.4-4 =0.4. \)

Hence,

\( \boxed{\mathrm{Var}(2-X)=0.4.} \)

Since adding or subtracting a constant changes only the mean (not the spread), the variance remains unchanged.

✅ Answer: \(\boxed{0.4}\)