Probabilityप्रायिकताquestions and answersप्रश्न और उत्तर
Probability questions in aptitude papers are counting questions in disguise. Fix the sample space correctly, count the favourable outcomes in that same space, divide, and reduce the fraction.
Unlimitedअसीमितfresh questionsनए प्रश्न
Answersउत्तरalways one tap awayहमेशा एक टैप दूर
1subtopicउप-टॉपिक
IBPS PO ~1
TCS NQT ~2
Practice questions are in English for now.अभ्यास के प्रश्न अभी अंग्रेज़ी में हैं।
How probability worksप्रायिकता को समझिए
Classical probability needs outcomes that are equally likely. With two dice the 36 ordered pairs, such as (2, 5) and (5, 2), are equally likely; the 11 possible sums are not. That is why a sum of 7 (6 ways) is six times as likely as a sum of 2 (1 way).
For “A or B”, add the two counts and subtract what was counted twice. In a pack of 52 cards there are 13 hearts and 4 kings, but the king of hearts is in both groups, so “a heart or a king” has 16 favourable cards, not 17. A standard pack has 4 suits of 13 cards and 12 face cards (jack, queen and king of each suit).
Without replacement, every draw changes the bag. Drawing two red balls from 5 red and 3 blue gives (5/8) × (4/7) = 5/14; the second fraction has one fewer red ball and one fewer ball in total. Counting pairs, C(5, 2)/C(8, 2) = 10/28, gives the same answer and is often quicker.
Each trick says when to use it, why it works and when it fails. Formula tricks are checked by code over their whole range.हर ट्रिक बताती है कि उसे कब इस्तेमाल करें, वह क्यों काम करती है और कब नहीं। सूत्र वाली ट्रिक्स को कोड से उनकी पूरी रेंज पर जाँचा गया है।
Mental mathsमानसिक गणितRarely taughtकम सिखाई जाने वाली✓ Checked on 11 cases✓ 11 मामलों पर जाँची गई
Ways to throw a sum s with two dice = 6 − |s − 7|
Use it when:कब इस्तेमाल करें: Any two-dice question about a particular sum, or a range of sums (“sum more than 8”, “sum is a prime”).
For a single sum s, the number of ordered ways is 6 − |s − 7| out of 36.
For a range, add the ways from the ladder 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 (sums 2 to 12).
Exampleउदाहरण
P(sum of two dice is 9)?
→ 6 − |9 − 7| = 4 ways, so 4/36 = 1/9.
Why it works, and when it failsयह क्यों काम करती है, और कब नहीं
Why:क्यों: For a sum s ≤ 7 the first die can show 1 to s − 1 (each fixing the second die), giving s − 1 ways; for s ≥ 7 it can show s − 6 to 6, giving 13 − s ways. Both are 6 − |s − 7|.
Limits:सीमाएँ: Only for two ordinary six-sided dice. For three dice the counts follow a different pattern (1, 3, 6, 10, 15, 21, 25, 27, 27, 25, …).
Option killerविकल्प-छँटनीRarely taughtकम सिखाई जाने वाली✓ Checked on 7,380 cases✓ 7,380 मामलों पर जाँची गई
Option killer: the reduced denominator must divide the sample space
Use it when:कब इस्तेमाल करें: A probability question with N equally likely outcomes (36 for two dice, 52 for a card, 2ⁿ for n coins) and fractional options.
Write the size of the sample space, N.
Reduce each option fully. Strike out any option whose denominator does not divide N.
Exampleउदाहरण
Two dice: P(sum is 7)? (a) 1/11 (b) 1/6 (c) 7/36 (d) 2/13
→ 11 and 13 do not divide 36, so (a) and (d) go at once. 7 ways would be too many (only 6 exist), so the answer is 1/6.
Why it works, and when it failsयह क्यों काम करती है, और कब नहीं
Why:क्यों: Any such probability is k/N for a whole number k. Reducing k/N divides N by gcd(k, N), so the reduced denominator is a factor of N.
Limits:सीमाएँ: Only for a single equally likely sample space. With two stages without replacement the space is ordered pairs (for two cards, 52 × 51), so use that product as N.
Formulaसूत्र✓ Checked on 195 cases✓ 195 मामलों पर जाँची गई
“At least one” = all − none
Use it when:कब इस्तेमाल करें: A committee, team or draw must contain “at least one” of some kind (a woman, a red ball, a senior member).
Count all selections with no restriction: C(n, k).
Count the selections with none of the required kind: C(others, k).
Subtract. For “at least two”, also subtract the selections with exactly one.
Exampleउदाहरण
A team of 3 from 5 men and 4 women with at least one woman.
→ C(9, 3) − C(5, 3) = 84 − 10 = 74.
Why it works, and when it failsयह क्यों काम करती है, और कब नहीं
Why:क्यों: Every selection either has no woman or has at least one, and never both, so the two counts add up to the total. Choosing “one woman, then anyone” instead counts a team with two women twice, once for each woman picked first.
Limits:सीमाएँ: Works whenever “none” is easy to count. If the condition is “at most” or “exactly”, count the allowed cases directly.
Solved examplesहल किए गए उदाहरण
Answers and full solutions are shown. Hide them to test yourself first.उत्तर और पूरे हल दिखाए गए हैं। पहले खुद को परखने के लिए इन्हें छिपाइए।
Example 1उदाहरण 1 · Probability with dice, coins, cards and bags · Easyआसान
During a Pongal card game in Ranchi, Isha draws one card at random from a well-shuffled pack of 52 cards. What is the probability that it is a queen or a club?
4/13
17/52
4/9
9/13
Answer:उत्तर:4/13
Favourable cards: 16.
Probability = 16/52 = 4/13.
Example 2उदाहरण 2 · Probability with dice, coins, cards and bags · Mediumमध्यम
Rohan rolls two dice for a board game. What is the probability that the two numbers differ by exactly 3?
1/12
1/7
1/6
5/6
Answer:उत्तर:1/6
Pairs (a, a + 3) with a from 1 to 3: 3. Reversed pairs: another 3.
Favourable: 6 of 36 → 1/6.
Example 3उदाहरण 3 · Probability with dice, coins, cards and bags · Hardकठिन
A bag in Fatima’s house in Nagpur has 4 red and 4 blue balls. Two are taken out at random, one after the other, without replacement. What is the probability that at least one of them is red?
3/14
3/4
11/14
23/28
Answer:उत्तर:11/14
P(no red) = C(4, 2)/C(8, 2) = 6/28.
P(at least one red) = 1 − 6/28 = 11/14.
Example 4उदाहरण 4 · Probability with dice, coins, cards and bags · Mediumमध्यम
Harpreet throws two fair dice together. What is the probability that the sum of the two numbers is 4?
1/18
1/12
1/11
11/12
Answer:उत्तर:1/12
Pairs adding to 4: (1, 3), (2, 2), (3, 1).
That is 3 of 36, so the probability is 1/12.
Example 5उदाहरण 5 · Probability with dice, coins, cards and bags · Mediumमध्यम
Two fair dice are thrown by Meera. What is the probability that the sum is greater than 10?
1/18
1/12
1/6
11/12
Answer:उत्तर:1/12
Ways for sums 11 to 12: 2 + 1 = 3.
Probability = 3/36 = 1/12.
Example 6उदाहरण 6 · Probability with dice, coins, cards and bags · Mediumमध्यम
Lakshmi rolls two dice at a mela in Guwahati. What is the probability that the two numbers differ by exactly 1?
5/36
5/21
5/18
13/18
Answer:उत्तर:5/18
Pairs (a, a + 1) with a from 1 to 5: 5. Reversed pairs: another 5.
Favourable: 10 of 36 → 5/18.
Common mistakesआम गलतियाँ
Treating unordered pairs or the 11 dice sums as equally likely instead of using the 36 ordered outcomes.
Using the with-replacement fraction (a/n)² when the question says the ball or card is not put back.
Giving the probability that the event does not happen, or the odds (favourable : unfavourable), instead of the probability.
Frequently asked questionsअक्सर पूछे जाने वाले प्रश्न
How many outcomes are there when two dice are thrown?
Thirty-six: each of the 6 faces of the first die can pair with each of the 6 faces of the second. Order matters, so (1, 6) and (6, 1) are different outcomes.
What is the difference between odds and probability?
Probability is favourable ÷ total. Odds in favour are favourable ÷ unfavourable. If 1 card in 4 is a heart, the probability is 1/4 but the odds in favour are 1 : 3.
When should I use the complement rule?
Whenever the question says “at least one” or the direct count has many cases. The chance of at least one head in 4 tosses is 1 − (1/2)⁴ = 15/16, since only one of the 16 outcomes has no head.