In another jackpot, Rohan also bought a ticket having a prize money of Rs 5,00,000. The chances of winning the jackpot are 1 in 1,00,000. Find the probability that on exactly one of tickets he wins the jackpot.

Probability

Probability of Exactly One of Two Independent Events

Official Answer

The probability that exactly one of Rohan's two jackpot tickets wins is $199998/10,000,000,000 \approx 0.00002$ (about $2 \times 10^{-5}$ ).

Given:

Probability of winning the first jackpot, $p_1 = 1/1,00,000$ (only one first-prize ticket among 1,00,000 tickets sold)
Probability of winning the second jackpot, $p_2 = 1/1,00,000$ (given directly)
The two tickets belong to two different, independent jackpots.

Required: exactly one of the two tickets wins — either the first ticket wins and the second loses, or the first loses and the second wins.

$P(\text{exactly one wins}) = p_1(1 - p_2) + (1 - p_1)p_2$

= $(1/100000)(99999/100000) + (99999/100000)(1/100000)$

$= 2 \times 99999/100000^2 = 199998/10,000,000,000 \approx 0.00002$

probabilityindependent eventsjackpotexactly one eventcomplementary probabilitymutually exclusive cases

Marking Scheme

11 mark: correct final probability $199998/10,000,000,000$ ( $\approx 0.00002$ ), obtained using the correct 'exactly one' formula $p_1(1-p_2)+(1-p_1)p_2$ with $p_1=p_2=1/1,00,000$ ; full marks only if the formula (not simple multiplication) is used.

Hint

Use $P(\text{exactly one}) = p_1(1-p_2) + (1-p_1)p_2$ with $p_1 = p_2 = 1/1,00,000$ — do not simply multiply the two probabilities together.

Quick Oral Answer

For two independent events A and B, the probability that exactly one occurs is $P(A)P(B') + P(A')P(B)$ ; here it works out to $2\times(1/100000)\times(99999/100000) \approx 0.00002$ .

Analysis & Explanation

This question tests the concept of finding the probability that exactly one of two independent events occurs — an extension of the addition and multiplication rules of probability.

Concept: For two independent events A and B with probabilities p₁ and p₂, 'exactly one occurs' means either (A occurs and B doesn't) or (A doesn't occur and B does). Since these two cases are mutually exclusive, their probabilities are added: $P(\text{exactly one}) = p_1(1-p_2) + (1-p_1)p_2$ . This differs from $P(A\cap B) = p_1p_2$ (both occur) and from $P(A\cup B) = p_1+p_2-p_1p_2$ (at least one occurs).

Exam trap: Students often confuse 'exactly one' with 'at least one'; the former excludes the case where both tickets win, while the latter includes it. Since p₁ and p₂ are both extremely small here, the numerical difference is tiny but the conceptual distinction is essential for full marks.

Real-world application: This exact reasoning underlies risk assessment in insurance (probability that exactly one of two independent policies results in a claim) and quality control (probability that exactly one of two independent machines produces a defective part).

Common Mistakes

1Multiplying the two probabilities directly $(p_1\times p_2)$ , which gives $P(\text{both win})$ , not $P(\text{exactly one wins})$ .
2Forgetting to add both complementary cases — (first wins, second loses) and (first loses, second wins) — and reporting only one term.
3Confusing 'probability of winning the jackpot' (top prize only, $1/1,00,000$ ) with 'probability of winning any prize' ( $6/1,00,000$ ) used in the earlier parts of this case study.

Interesting Facts

The formula for 'exactly one of two independent events' — $P(A)P(B')+P(A')P(B)$ — is a special case of the inclusion–exclusion principle used throughout risk modelling, insurance, and gaming mathematics.

Indian lottery and online-jackpot regulations require operators to publish the exact odds of winning — exactly the kind of number (1 in 1,00,000) used in this problem.

Even though winning a single jackpot is rare (probability 0.00001), holding tickets in two independent jackpots only raises the chance of winning at least one to about 0.00002 — a real illustration of why buying a few extra lottery tickets barely moves the odds.

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Frequently Asked Questions

Why isn't the answer simply $p_1 \times p_2$ ?

$p_1\times p_2$ gives the probability that BOTH tickets win, not that exactly one wins. 'Exactly one' requires adding the two mutually exclusive cases where one ticket wins and the other loses.

Are the two jackpots independent events?

Yes — the outcome of one online jackpot draw has no bearing on the outcome of the other, so the multiplication rule for independent events applies directly.