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- Question
- 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
- Correct Answer
- 2/7
ExplanationA 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 - 1. 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
- 2. 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
- 3. 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 Discuss
- 4. 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
- 5. All the boys are sitting together.
Options- A. 3/42
- B. 41/42
- C. 3/35
- D. 1/42 Discuss
- 6. 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
- 7. 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
- 8. A natural number is selected at random from the set x = { x : 1 ? x ? 100 }. The probability that the number satisfies the inequality x 2 - 13x ? 30 is?
Options- A. 5/9
- B. 9/50
- C. 3/20
- D. 7/9 Discuss
- 9. The probability that 9 students pass in Mathematics, Physics and Chemistry are m, p and c respectively, of these subject the students has a 75% chance of passing in at least one subject 50% chance of passing in at least two subject and 40% chance of passing in exactly two. Which of the following relation are true?
Options- A. p + m + c =19/20
- B. p + m + c = 17/20
- C. pmc = 1/10
- D. pmc = 1/4 Discuss
- 10. If the integer m and n are chosen at random between 1 and 100, then the probability that a number of the from 7 m + 7 n is divisible by 5 equals?
Options- A. 1/4
- B. 1/7
- C. 1/8
- D. 1/49 Discuss
Probability problems
Search Results
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: 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: 11/13
Explanation:
Total 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
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: 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
Correct Answer: 3/20
Explanation:
x2 - 13 x - 30 ? 0
? (x + 2) (x - 15) ? 0
? -2 ? x ? 15
But x is a natural number.
? 1 ? x ? 15
? Reqd. Probability P = 15/100 = 3/20
Correct Answer: pmc = 1/10
Explanation:
P(M) = m, P (p) = p, P(c) = c
? The probability of at least one success
= P (M ? P ? C)
= m + p + c -mp - mc - pc + mcp = 3/4 ...(1)
The probability of at least two successes = mcp + mcp + mcp + mcp
= mc(1 - p) + mp (1 - c ) + (1 - m )cp + mcp
= mc + mp + cp - 2mcp = 1/2
The probability of exactly two success
= mcp + mcp + mcp
= mc(1 - p) + mp (1 - c ) cp(1 - m )
= mc + mp + cp - 3 mcp = 2/5
(2) & (3) gives,
? mcp = 1/2 - 2/5 = 1/10
? mc + mp + cp = 2/10 + 1/2 = 1/5 + 1/2 = 7/10
From (1),
m + p + c - 7/10 + 1/10 = 3/4
? m + p + c = 3/4 + 7/10 - 1/10 = 27/20
Thus, pmc = 1/10 is a true relation.
Correct Answer: 1/8
Explanation:
Number of possible cases = 100 x 100 for 7m + 7n to be divisible by 5,
One of the term has to end with 9 and other with 1 (7m cannot be divided by 9 )
? m can be 2, 6 , 10, 14, ..... 98 (25 values) and n can be 4, 8, 12 ...... 100(25 values)
Since m and n can interchange
? Required probability = (2 x 25 x 25) / (100 x 100) = 1/8
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