Modern Math - Probability - Previous Year IPM/BBA Questions
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The Probability that a randomly chosen positive divisor of 102023 is an integer multiple of 102001 is
- (a)
23 / 2024
- (b)
232 / 20242
- (c)
223 / 20232
- (d)
22 / 2023
Answer: Option B
Text Explanation :
Workspace:
If a three-digit number is chosen at random, what is the probability that it is divisible neither by 3 nor by 4?
- (a)
1/2
- (b)
1/4
- (c)
2/3
- (d)
1/3
Answer: Option A
Text Explanation :
Workspace:
The minimum number of times a fair coin must be tossed so that probability of getting at least one head exceed 0.8 is
- (a)
3
- (b)
7
- (c)
5
- (d)
6
Answer: Option A
Text Explanation :
Workspace:
If one of the factors of the number 3728173 is randomly chosen, then the probability that the chosen factor will be a perfect square is
- (a)
5/36
- (b)
1/12
- (c)
3/40
- (d)
5/32
Answer: Option A
Text Explanation :
Number of factors of 3728173 = (7 + 1)(8 + 1)(3 + 1) = 8 × 9 × 4 = 288
Number of factors of 3728173 which are perfect squares = (4)(5)(2) = 40
Required probability = 40/288 = 10/72 = 5/36.
Hence, option (a).
Workspace:
The probability that a randomly chosen factor of 1019 is a multiple of 1015 is
- (a)
1/25
- (b)
1/12
- (c)
1/20
- (d)
1/16
Answer: Option D
Text Explanation :
Total number of factors of 1019 = 219 × 519 are (19 + 1) × (19 + 1) = 400
Factors of 219 × 519 which are also a multiple of 1015 will have power of both 2 and 5 greater than or equal to 15.
Possible powers of 2 or 5 are 15 or 16 or 17 or 18 or 19 i.e., 5 possibilities.
∴ Number of such factors are 5 × 5 = 25
⇒ Required probability = 25/400 = 1/16
Hence, option (d).
Workspace:
A man is known to speak the truth on an average 4 out of 5 times. He throws a die and reports that it is a five. The probability that it is actually a five is
- (a)
4/9
- (b)
5/9
- (c)
4/15
- (d)
2/15
Answer: Option E
Text Explanation :
The answer should be 4/5.
P(A given B) = P(A and B) ÷ P(B)
P(5 given that he says it is a 5) = P(5 and he says it is 5) ÷ P(he says it is 5)
P(5 and he says it is 5) = P(it is 5) × P(he say truth) = 1/6 × 4/5 = 4/30
P(he says it is 5) = P(it is 5) × P(he says the truth) + P(it is not 5) × P(he lies) × P(he lies that it is 5) = 1/6 × 4/5 + 5/6 × 1/5 × 1/5 = 4/30 + 1/30 = 5/30
∴ P(5 given that he says it is a 5) = 4/30 ÷ 5/30 = 4/5.
Hence, 4/5.
Note:
Let's say the die rolls a 4 and the person lies, the person could lie that it is a 1 or 2 or 3 or 5 or 6.
Now, probability that the person lies that is it 5 = 1/5
∴ P(it is not 5) × P(he lies) × P(he lies that it is 5) = 5/6 × 1/5 × 1/5 = 1/30.
In most study material you would find the answer to this question as 4/9.
This is because they assume that when the die does not roll a 5 and he lies, he will always say that it is 5. But this is not true. He could lie that it is a 4 when a 6 rolls. Hence, the difference in the answer.
Workspace:
From a pack of 52 cards, we draw one by one, without replacement. If f(n) is the probability that an Ace will appear at the nth turn, then
- (a)
f(2) = 1/13 > f(3)
- (b)
1/13 > f(2) > f(3)
- (c)
f(3) > f(2) = 1/13
- (d)
f(2) = f(3) = 1/13
Answer: Option D
Text Explanation :
Workspace:
A die is thrown three times and the sum of the three numbers is found to be 15. The probability that the first throw was a four is
- (a)
1/6
- (b)
1/4
- (c)
1/5
- (d)
1/10
Answer: Option C
Text Explanation :
Workspace: