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?

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

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

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

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

n(S) = 2¹⁰⁰
n(E) = No. of favourable ways = ¹⁰⁰C₁ + ¹⁰⁰C₃ + ... ¹⁰⁰ C₉₉ = 2^100-1 = 2⁹⁹
[? ⁿC₁ + ⁿC₃ + ⁿC₅ + ........................... = 2^n-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. "

The total number of cause is 2¹⁰⁰.
The number of favourable cases are ¹⁰⁰C₁ + ¹⁰⁰C₃ + ..... + ¹⁰⁰C₉₉
= 2^{100 - 1} = 2⁹⁹
? Reqd probability = 2⁹⁹/ 2¹⁰⁰ = 1/2

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

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.

Number of possible cases = 100 x 100 for 7^m + 7ⁿ to be divisible by 5,
One of the term has to end with 9 and other with 1 (7^m 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

If the last digit in the product is to 2, 4, 6, 8 the last digit in all the n number should not be 0 and 5 and the last digit of all number should not be selected exclusively from the set of number {1, 3, 7, 9}
? Favourable number of cases
= 8ⁿ - 4ⁿ
But generally the last digit can be one of 0, 1, 2, 3, .... 9.
Hence, the total number of ways = 10ⁿ
Hence, the required probability
= 8ⁿ - 4ⁿ / 10ⁿ
= 4ⁿ - 2ⁿ / 5ⁿ