Probability
- A fair coin is tossed 100 times, The probability of getting head an odd number of times is ?
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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 = 1/2Correct Option: C
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. "
- 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 ?
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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.Correct Option: B
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
- 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 ?
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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 = 54/55Correct Option: B
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
- What is the probability that a leap year selected randomly will have 53 Mondays ?
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A leap year has 366 days = 52 weeks + 2 days.
Correct Option: A
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
- What is the probability that a leap year selected randomly will have 53 Mondays ?
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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)Correct Option: A
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