JEE MAIN Probability PNC Previous Year Questions (PYQs) – Page 1 of 2
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Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is :
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Solution
Let the 11 consecutive natural numbers be
$1, 2, 3, \dots, 11.$
Total ways to choose any 3 numbers = $\displaystyle \binom{11}{3} = 165.$
Now, we need to count the number of 3-number selections that can form an arithmetic progression (A.P.) with positive common difference.
For an A.P., let the middle term be $a$ and common difference be $d>0$. Then the three terms are: $(a-d,\ a,\ a+d)$ These must all lie between $1$ and $11$.
That means $1 \le a-d$ and $a+d \le 11$ ⟹ $d \le \min(a-1,\ 11-a)$
Now we count possible values of $d$ for each $a$:
Total = $1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25.$
Hence, number of favorable triplets = $25.$
Therefore, $\displaystyle P = \frac{25}{165} = \frac{5}{33}.$
Final Answer: $\boxed{\dfrac{5}{33}}$
Now, we need to count the number of 3-number selections that can form an arithmetic progression (A.P.) with positive common difference.
For an A.P., let the middle term be $a$ and common difference be $d>0$. Then the three terms are: $(a-d,\ a,\ a+d)$ These must all lie between $1$ and $11$.
That means $1 \le a-d$ and $a+d \le 11$ ⟹ $d \le \min(a-1,\ 11-a)$
Now we count possible values of $d$ for each $a$:
| $a$ | $\min(a-1,\ 11-a)$ | Possible $d$ values |
|---|---|---|
| 1 | 0 | – |
| 2 | 1 | 1 |
| 3 | 2 | 1,2 |
| 4 | 3 | 1,2,3 |
| 5 | 4 | 1,2,3,4 |
| 6 | 5 | 1,2,3,4,5 |
| 7 | 4 | 1,2,3,4 |
| 8 | 3 | 1,2,3 |
| 9 | 2 | 1,2 |
| 10 | 1 | 1 |
| 11 | 0 | – |
Hence, number of favorable triplets = $25.$
Therefore, $\displaystyle P = \frac{25}{165} = \frac{5}{33}.$
Final Answer: $\boxed{\dfrac{5}{33}}$
An urn contains $6$ white and $9$ black balls. Two successive draws of $4$ balls are made without replacement. The probability that the first draw gives all white balls and the second draw gives all black balls is:
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Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is :
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Solution
In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is :
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Let E1 and E2 be two events such that the conditional probabilities $P({E_1}|{E_2}) = {1 \over 2}$, $P({E_2}|{E_1}) = {3 \over 4}$ and $P({E_1} \cap {E_2}) = {1 \over 8}$. Then :
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Solution
A fair die is thrown until $2$ appears. Then the probability that $2$ appears in an even number of throws is:
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Solution
A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is :
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A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :
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Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to :
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Solution
Two dies are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :
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