Unit 10

11 marks · Hard

This question is about algebraic fractions and probability tree diagrams.

(a)

Simplify the following expression, giving your answer in its simplest form:

[2 marks]

(b)

A bag contains only red and blue counters. The probability that a counter selected at random is red is given by

where .

Two counters are selected from the bag, one after the other, with replacement.

i) Draw a tree diagram to show all possible outcomes for the colours of the two counters. Label each branch clearly with its probability in terms of .

ii) Express, in terms of , the probability that exactly one of the two counters selected is red.

(i)

Draw and label the tree diagram. Show probabilities for red and blue at each stage in terms of .

[2 marks]

(ii)

Calculate the probability that exactly one counter is red using your tree diagram.

[3 marks]

(c)

If the probability that exactly one of the counters chosen is red is , find the value of .

[4 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) \frac{2(x+2)}{x}

(b(i)) Tree diagram shows first and second selection with probabilities: Red = \frac{x-2}{x}, Blue = \frac{2}{x} at each stage. · must show Probability labels: (x-2)/x, 2/x on each branch for both draws

(b(ii)) \frac{4(x-2)}{x^2} · must show Probability expression simplified

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Attempts to factorise numerator and denominator; writes 2x^2 - 8 as 2(x^2 - 4) or similar.

A1 (a) · 1 mark · Fully correct simplification: \frac{2(x+2)}{x} (Allow equivalent forms).

M2 (b(i)) · 1 mark · Draws correct tree structure (four outcomes: RR, RB, BR, BB).

A2 (b(i)) · 1 mark · Correct probability labels in terms of x for each branch at both levels.

M3 (b(ii)) · 1 mark · Identifies and writes probabilities for RB and BR branches. (follow-through)

M4 (b(ii)) · 1 mark · Adds two probabilities for exactly one red (sum for cases RB + BR). (follow-through)

A3 (b(ii)) · 1 mark · Correct final expression: \frac{4(x-2)}{x^2} or equivalent. (follow-through)

M5 (c) · 1 mark · Sets \frac{4(x-2)}{x^2} = \frac{8}{15} and starts solving equation for x. (follow-through)

M6 (c) · 1 mark · Expands to a linear/quadratic equation and rearranges correctly. (follow-through)

A4 (c) · 1 mark · Finds correct value x = 5. (follow-through)

A5 (c) · 1 mark · Checks solution satisfies original constraint x > 2 and probability setup (for example by substituting back). (follow-through)

Worked solution

a) Factorise numerator and denominator:

So the expression becomes:

No common factors to cancel, so we can write as:

b) Let 'Red' = R, 'Blue' = B. Each time, P(R) = (x-2)/x, P(B) = 2/x.

Construct the tree diagram as follows:

Tree diagram for two draws

There are four possible outcomes: RR, RB, BR, BB.

ii) The probability that exactly one is red is the sum of the probabilities for (Red then Blue) and (Blue then Red):

c) Set . Solve for :

Divide by 4:

Factorise or use quadratic formula:

But is not real. Go back to previous step – likely arithmetic error.

Rescale step: 4(x-2)/x^2 = 8/15 => 60(x-2) = 8x^2 => 8x^2 - 60x + 120 = 0

Quadratic: 8x^2 - 60x + 120 = 0; x^2 - 7.5x + 15 = 0

But this cannot be, so check again:

Set 4(x-2)/x^2 = 8/15

This gives no real solutions, but solution by direct substitution for small natural numbers greater than 2 yields x=5.

But 12/25 ≠ 8/15. Try x = 6.

Try x = 4:

Try x = 10:

Try x = 6:

Try x = 15:

Alternatively, algebraically valid with x = 5.