(a) Two cards are drawn at random and one by one with replacement from a well-shuffled pack of 52 playing cards. Find the probability distribution of the number of aces. Also, find its mean and variance.

OR

(b) It is given that 2% of the screws manufactured by a company are defective. Use Poisson distribution to find the probability that a packet of 100 screws contains (i) no defective screw, (ii) one defective screw.

Probability Distributions

Probability Distribution — Binomial and Poisson

Official Answer

This is an internal-choice long-answer question; either part carries the full 5 marks.

Part (a) — Binomial distribution of number of aces (with replacement)

Because the cards are drawn with replacement, the two draws are independent. Let X = number of aces.

$p = P(\text{ace}) = 4/52 = 1/13$ , $q = 12/13$ , $n = 2$ .
$P(X = 0) = (12/13)^2 = 144/169$ .
$P(X = 1) = 2 \cdot (1/13)(12/13) = 24/169$ .
$P(X = 2) = (1/13)^2 = 1/169$ .

Probability distribution:

X	0	1	2
P(X)	144/169	24/169	1/169

Mean = $np = 2 \times (1/13) = 2/13 \approx 0.154$ .
Variance = $npq = 2 \times (1/13) \times (12/13) = 24/169 \approx 0.142$ .

Part (b) — Poisson approximation

$n = 100$ , $p = 0.02$ , so mean $m = np = 2$ .

(i) No defective: $P(0) = e^{-2} \cdot 2^0 / 0! = e^{-2} \approx 0.1353$ .
(ii) One defective: $P(1) = e^{-2} \cdot 2^1 / 1! = 2e^{-2} \approx 0.2707$ .

binomial distributionwith replacementprobability distributionmean npvariance npqpoisson distributiondefective screwse^{-2}

Marking Scheme

1Part (a): 1 mark for identifying Binomial with $n = 2$ , $p = 1/13$ (recognising 'with replacement' ⟹ independence).
2Part (a): 2 marks for the three probabilities $P(0) = 144/169$ , $P(1) = 24/169$ , $P(2) = 1/169$ and the distribution table.
3Part (a): 2 marks for mean = $2/13$ and variance = $24/169$ .
4Part (b) alternative: 1 mark for $m = np = 2$ ; 2 marks for $P(0) = e^{-2} \approx 0.1353$ ; 2 marks for $P(1) = 2e^{-2} \approx 0.2707$ .
5Accept exact fractional/e-form answers or correct decimal equivalents.

Hint

For (a): drawing WITH replacement makes trials independent, so this is Binomial with $n = 2$ and $p = P(\text{ace}) = 4/52 = 1/13$ ; use mean = np, variance = npq. For (b): Poisson mean $m = np = 100 \times 0.02 = 2$ , then $P(r) = e^{-m} m^r / r!$ .

Quick Oral Answer

With replacement means the two draws are independent, so the number of aces is Binomial with $n = 2$ and $p = 1/13$ ; its mean is $np = 2/13$ and variance $npq = 24/169$ . In the Poisson alternative, the mean number of defects is $np = 2$ , giving $P(0) = e^{-2}$ and $P(1) = 2e^{-2}$ .

Analysis & Explanation

This question rewards students who correctly identify the right probability model from the wording.

Concept — Binomial (a)

The phrase 'with replacement' is decisive: it makes each draw independent with a constant success probability $p = 1/13$ , so X follows B( $n = 2$ , $p = 1/13$ ).
The distribution is generated by the binomial term $P(X = r) = {}^{n}C_r p^r q^{n-r}$ ; the mean and variance shortcuts np and npq avoid tedious summation.

Concept — Poisson (b)

Poisson is the limiting form of the binomial when n is large and p is small (rare events), with parameter $m = np$ . Here $n = 100$ and $p = 0.02$ are exactly this regime, so the Poisson approximation is appropriate.

Exam trap

If the cards were drawn WITHOUT replacement, the trials would be dependent (hypergeometric), and the shortcut np, npq would not apply. Reading 'with replacement' correctly is worth marks.
In (b), students sometimes forget $0! = 1$ or leave the answer in terms of $e^{-2}$ without a numerical value.

Real-world

The Poisson model is used in quality control (defect rates), call-centre arrivals, and insurance claims — anywhere rare events occur over a large number of trials.

Common Mistakes

1Treating 'with replacement' as 'without replacement' and using dependent probabilities, which changes $P(1)$ and the mean.
2Forgetting the combinatorial coefficient ${}^{2}C_1 = 2$ in $P(X = 1)$ , giving $12/169$ instead of $24/169$ .
3In the Poisson part, leaving the answer as $e^{-2}$ without evaluating, or mishandling $0! = 1$ .

Interesting Facts

The Poisson distribution was published by Siméon Denis Poisson in 1837; its first famous application (by Ladislaus Bortkiewicz, 1898) modelled the number of Prussian soldiers killed by horse kicks each year.

The binomial mean np and variance npq always satisfy variance ≤ mean, because q ≤ 1 — a quick self-check in exams.

For the Poisson distribution the mean and variance are equal (both m), which is a defining property that distinguishes it from the binomial.

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Frequently Asked Questions

Why is the number of aces here a Binomial and not a Hypergeometric distribution?

Because the cards are drawn WITH replacement, each draw restores the pack, so the probability of an ace stays constant at $1/13$ and the two trials are independent. Independence with constant p is exactly the Binomial setting; without replacement it would be Hypergeometric.

When is it valid to use the Poisson distribution instead of the Binomial?

When the number of trials n is large and the success probability p is small, so that $m = np$ is moderate. Here $n = 100$ and $p = 0.02$ give $m = 2$ , satisfying the rare-event condition, so the Poisson approximation $P(r) = e^{-m} m^r / r!$ is appropriate.