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Discrete Random Variables: Question 7

Syllabus 5.4

Structured AS 8 marks

In a large batch of memory cards, the probability that a randomly chosen card is defective is 0.150.15, independently of any other card. A quality inspector selects cards one at a time, at random from the batch, and tests them until she finds the first defective card. Let XX be the number of cards tested up to and including the first defective card found, so that X∼Geo(0.15)X\sim \mathrm{Geo}(0.15).

(a) State two conditions, in the context of this situation, that must hold for XX to be modelled by the geometric distribution Geo(p)\mathrm{Geo}(p). [2]

(b) Find P(X=5)P(X=5). [2]

(c) Find P(X≤3)P(X \leq 3). [3]

(d) State E(X)E(X). [1]

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Worked solution

Part (a): Conditions for a geometric model

For XX to be modelled by a geometric distribution Geo(p)\mathrm{Geo}(p), the situation must satisfy:

  1. Each card tested has exactly two possible outcomes, “defective” or “not defective”.
  2. The probability that a card is defective is constant for every card, here p=0.15p=0.15.
  3. The cards are independent of one another.
  4. Testing continues one at a time until the first success (the first defective card) occurs.

Any two of these conditions is sufficient for the marks, e.g. “the probability that a card is defective is constant at 0.150.15 for every card, and whether one card is defective is independent of any other card.”

Since these hold, X∼Geo(0.15)X\sim\mathrm{Geo}(0.15).

Part (b): Finding P(X = 5)

Using P(X=r)=p(1−p)r−1P(X=r)=p(1-p)^{r-1} with p=0.15p=0.15, q=1−p=0.85q=1-p=0.85, r=5r=5, the first four cards tested must all be non-defective, and the fifth must be defective: P(X=5)=0.15×(0.85)4P(X=5)=0.15\times(0.85)^{4}

Compute (0.85)4(0.85)^4 step by step: (0.85)2=0.7225(0.85)^2=0.7225 (0.85)4=(0.7225)2=0.52200625(0.85)^4=(0.7225)^2=0.52200625

So P(X=5)=0.15×0.52200625≈0.078301P(X=5)=0.15\times0.52200625\approx0.078301

P(X=5)≈0.0783P(X=5)\approx\boxed{0.0783} (to 3 significant figures).

Part (c): Finding P(X ≤ 3)

"X≤3X\leq3" means the first defective card is found on the 1st, 2nd, or 3rd test. Equivalently, the complement of “all of the first 3 cards are non-defective”: P(X≤3)=1−q3=1−(0.85)3P(X\leq3)=1-q^{3}=1-(0.85)^{3}

Compute (0.85)3(0.85)^3: (0.85)3=(0.85)2×0.85=0.7225×0.85=0.614125(0.85)^3=(0.85)^2\times0.85=0.7225\times0.85=0.614125

So P(X≤3)=1−0.614125=0.385875P(X\leq3)=1-0.614125=0.385875

P(X≤3)≈0.386P(X\leq3)\approx\boxed{0.386} (to 3 significant figures).

Check by direct summation: P(X=1)+P(X=2)+P(X=3)=0.15+(0.85)(0.15)+(0.85)2(0.15)P(X=1)+P(X=2)+P(X=3)=0.15+(0.85)(0.15)+(0.85)^2(0.15) =0.15+0.1275+0.108375=0.385875=0.15+0.1275+0.108375=0.385875 which agrees exactly with the complement calculation.

Part (d): Finding E(X)

For a geometric distribution, E(X)=1pE(X)=\dfrac{1}{p}: E(X)=10.15=6.66‾≈6.67E(X)=\frac{1}{0.15}=6.6\overline{6}\approx6.67

Final answers

  • (a) Any two valid conditions, e.g. constant probability p=0.15p=0.15 and independence between cards.
  • (b) P(X=5)≈0.0783P(X=5)\approx\boxed{0.0783}
  • (c) P(X≤3)≈0.386P(X\leq3)\approx\boxed{0.386}
  • (d) E(X)=10.15≈6.67E(X)=\dfrac{1}{0.15}\approx\boxed{6.67}