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- Question
- Out of 15 students studying in a class, 7 are from Maharastra, 5 are from Karnataka and 3 are from Goa, Four students are to be selected at random. What are the chances that at least one is from Karnataka?
Options- A. 12/13
- B. 11/13
- C. 10/15
- D. 1/15
- Correct Answer
- 11/13
ExplanationTotal possible ways of selecting 4 students out of 15 students = 15C4
= (15 x 14 x 13 x 12) / (1 x 2 x 3 x 4) = 1365
The no. of ways of selecting 4 students in which no students belongs to karnataka = 10C4
? Number of ways of selecting atleast one student from karnataka = 15C4 - 10C4 = 1155.
? Required probability
= 1155 / 1365 = 77 / 91 = 11 / 13 - 1. A fair coin is tossed 100 times, The probability of getting head an odd number of times is?
Options- A. 1/4
- B. 2/3
- C. 1/2
- D. 3/4 Discuss
- 2. All the boys are sitting together.
Options- A. 3/42
- B. 41/42
- C. 3/35
- D. 1/42 Discuss
- 3. All the girls are sitting together.
Options- A. 1/42
- B. 41/42
- C. 3/35
- D. 1/42 Discuss
- 4. All the boys are not sitting together?
Options- A. 3/35
- B. 43/49
- C. 41/42
- D. 1/42 Discuss
- 5. All the girls are not sitting together?
Options- A. 41/42
- B. 39/42
- C. 31/42
- D. 31/35 Discuss
- 6. In a box carrying one dozen of oranges, one third have become bad. If 3 oranges are taken out from the box at random, what is the probability that at least one oranges out of the three oranges picked up is good?
Options- A. 1/55
- B. 54/55
- C. 45/55
- D. 3/55 Discuss
- 7. What is the probability that a leap year selected randomly will have 53 Mondays?
Options- A. 2/7
- B. 5/7
- C. 1/7
- D. Data inadequate Discuss
- 8. What is the probability that a leap year selected randomly will have 53 Mondays?
Options- A. 2/7
- B. 5/7
- C. 1/7
- D. Data inadequate Discuss
- 9. If the probability for A fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is?
Options- A. 0.38
- B. 0.44
- C. 0.50
- D. 0.94 Discuss
- 10. A fair coin is tossed 100 times . The probability of getting tails an odd number of times is?
Options- A. 1/2
- B. 1/8
- C. 3/8
- D. None of these Discuss
Probability problems
Search Results
Correct Answer: 1/2
Explanation:
n(S) = 2100
n(E) = No. of favourable ways = 100C1 + 100C3 + ... 100 C99 = 2100-1 = 299
[? nC1 + nC3 + nC5 + ........................... = 2n-1 ]
? P(E)= n(E)/n(S) = 299/2100 = 1/2
Note : The given case can be generalised as "If a unbiased coin is tossed 'n' times, then the chance that the head will present itself an odd number of times is 1/2. "
Correct Answer: 1/42
Explanation:
Required probability = 6!5!/10! = 1/42
Correct Answer: 1/42
Explanation:
We have, 5 boys and 5 girls.
Since, all the girls are sitting together, we consider them as one. Now, we can arrange 5 boys and 1 girl in 6! ways and these 5 girls (whom we considered as one) can also be arranged in 5! ways.
? Favorable number of ways = 6!5!
and total number of ways to arrange 5 boys and 5 girls = 10!
? Required probability 6!5!/10! = 1/42
Correct Answer: 41/42
Explanation:
Required probability = 1- probability that all the boys sit together
= 1- 1/42 = 41/42
Correct Answer: 41/42
Explanation:
Required probability = 1- probability that all the girls sit together
= 1 - 1/42 = 41/42
Correct Answer: 54/55
Explanation:
n(S) = 12C3 = (12 x 11 x 10) / (3 x 2) = 2 x 11 x 10 = 220
No. of selecting of 3 oranges out of the total 12 orange = 4C3 = 4
? n(E) = No. of desired selection of oranges = 220 - 4 = 216
? P(E) = n(E) / n(S) = 216 / 220 = 54/55
Correct Answer: 2/7
Explanation:
A leap year has 366 days = 52 weeks + 2 days. These 2 days can be (Sunday, Monday, Wednesday) .... or (Saturday, Sunday). Out of these total 7 out comes there are 2 cases favourable to the desired event i, e. (Sunday, Monday) and (Monday, Tuesday)
? Required probability = 2/7
Correct Answer: 2/7
Explanation:
A leap year has 366 days = 52 weeks + 2 days.
The extra days 2 days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday) .... or (Saturday, Sunday).
Out of these total 7 out comes there are 2 cases favorable to the desired event i,e. (Sunday, Monday) and (Monday, Tuesday)
? Required probability = 2/7
Correct Answer: 0.44
Explanation:
? P(A) = 0.2
? P(A) = 1 - 0.0 = 0.8
And P (B) = 0.3
? P(B) = 1 - 0.3 = 0.7
Required probability
= P(A ? B)
= P(A ? B)
= 1 - P (A ? B ) = 1 - P(A) P(B)
= 1 - (0.8) (0.7) = 1-0.56 = 0.44
Correct Answer: 1/2
Explanation:
The total number of cause is 2100.
The number of favourable cases are 100C1 + 100C3 + ..... + 100C99
= 2100 - 1 = 299
? Reqd probability = 299/ 2100 = 1/2
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