Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Let a, b, $c \in \{1, 2, 3, 4\}$. If the probability, that $ax^2 + 2\sqrt{2}bx + c > 0$ for all $x \in R$, is $\frac{m}{n}$, $\gcd(m, n) = 1$, then $m+n$ is equal to _________.
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Quadratic inequalities and probability - condition for quadratic to be always positive using discriminant
A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is $\frac{m}{n}$, gcd $(m, n) = 1$, then $m + n$ is equal to:
(1) 53
(2) 55
(3) 107
(4) 105
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Answer: Option (3)
Key idea: Probability with constraints - finding all possible outcomes satisfying a linear equation and computing total probability
Set up equation: $10H + 5T = 30$ where $H$ = heads, $T$ = tails. Simplify to $T = 6 - 2H$
A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is $p$, then $96p$ is equal to _____.
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Answer: 9
Key idea: Conditional Probability with Overlapping Events
Identify overlapping tosses: First 6 tosses are positions 1-6, last 5 tosses are positions 4-8. Positions 4, 5, 6 overlap.
Set up equations: Let $y$ = heads in positions 4-6. Then heads in positions 1-3 is $x = 4-y$ and heads in positions 7-8 is $z = 3-y$.
Find valid range: From constraints $x, z \geq 0$ and $x \leq 3, y \leq 3, z \leq 2$, we get $1 \leq y \leq 3$.
Count favorable outcomes: For $y=1$: $C(3,3) \cdot C(3,1) \cdot C(2,2) = 3$; For $y=2$: $C(3,2) \cdot C(3,2) \cdot C(2,1) = 18$; For $y=3$: $C(3,1) \cdot C(3,3) \cdot C(2,0) = 3$. Total = 24.
From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to $\frac{a}{b}$, where $a, b \in \mathbb{N}$ and $gcd(a, b) = 1$, then $a + b$ is equal to
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Answer: 944
Key idea: Counting favorable outcomes in arithmetic progressions and probability
Total ways to select 3 dates from 31: $\binom{31}{3} = \frac{31 \times 30 \times 29}{6} = 4495$
For dates in increasing A.P.: Let them be $a, a+d, a+2d$ where $1 \leq a \leq 31-2d$ and $d \geq 1$
Maximum $d = 15$ (when $a=1, a+2d=31$). For each $d$, number of A.P.s = $31-2d$
A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:
(1) $\frac{7}{10}$
(2) $\frac{10}{17}$
(3) $\frac{12}{19}$
(4) $\frac{7}{19}$
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Answer: Option (2)
Key idea: Bayes' Theorem and Conditional Probability
Identify consecutive letter pairs: KANPUR has 5 pairs (KA, AN, NP, PU, UR) with 1 'AN', ANANTPUR has 7 pairs (AN, NA, AN, NT, TP, PU, UR) with 2 'AN's
The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is:
A bag contains $(N + 1)$ coins $-$ N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then N is equal to:
Apply law of total probability: $P(\text{Head}) = P(\text{fair coin}) \times P(H|\text{fair}) + P(\text{double-headed}) \times P(H|\text{double-headed})$
A bag contains $6$ blue and $6$ green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is :
(1) $\frac{63}{925}$
(2) $\frac{17}{231}$
(3) $\frac{16}{231}$
(4) $\frac{64}{925}$
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Answer: Option (3)
Key idea: Probability with combinations and arrangements
We have 6 blue and 6 green balls. We draw 6 pairs without replacement.
**Total arrangements**: Arrange 6 B's and 6 G's in 12 positions where positions form pairs: $(1,2), (3,4), (5,6), (7,8), (9,10), (11,12)$. Total = $\binom{12}{6} = 924$
**Favorable arrangements**: Each pair must have exactly 1 blue and 1 green ball.
For each of the 6 pairs, choose which position gets blue (other gets green): $2^6 = 64$ ways
A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are $\frac{2}{5}$, $\frac{1}{5}$ and $\frac{2}{5}$. The probabilities that the candidate reaches late at the examination centre are $\frac{1}{5}$, $\frac{1}{3}$ and $\frac{1}{4}$ if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is:
(1) $\frac{11}{37}$
(2) $\frac{12}{37}$
(3) $\frac{13}{37}$
(4) $\frac{14}{37}$
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Answer: Option (2)
Key idea: Bayes' Theorem and Law of Total Probability
