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?

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.

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

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

? 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

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

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

The probability that head is show in one coin is 1/2.

The probability that the sum of the number on the dice is a prime = the probability that the following pair of number on the dice is a getting on the dice, namely (1, 1), (1, 2), (2, 1), (1, 4 ), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5), (5, 6) = 15/36.

? The required probability = 1/2 x 1/2 x 15/36 = 5/48.

Let A = Event that A speaks the truth
and B = Event that B speaks the truth
Then, P(A) = 60/100 = 3/5 and P(A) = 80/100 = 4/5
? P(A) = (1 - 3/5) = 2/5 and
P(B) = (1 - 4/5) = 1/5

P(A and B contradict each other)
= P [(A speaks the truth and B tells a lie) or (A tells a lie and B speak the truth)]
=P[(A and B) or (A and B)]
=P(A and B) + P (A and B)
= P(A) x P(B) + P(A) x P(B)
= (3/5 x 1/5) + ( 2/5 x 4/5)
= (3/25 + 8/25 ) = 11/25
= (11/25 x 100)% = 44%

In a leap year, there are 366 days. It means 52 full weeks + 2 odd days. These two days can be (Mon, Tues), (Tues, Wed), (Wed, Thurs), (Thurs, Fri), (Fri, Sat), (Sat, Sun) or (Sun, Mon),
So, Required probability = 2/7

Total number of cases = ¹⁰c₄
Favorable number of cases = ⁴c₂. ⁶c₂
{Since, we are to select 2 children out of 4 and remaining 2 persons are to be selected from remaining 6 persons ( 2W + 4M)}
? required Probability = ⁴c₂ . ⁶c₂ / ¹⁰c₄
= [{(4 x 3) / (2 x 1)} x {(6 x 5) / (2 x 1)}] / [ (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)]
= [(12/2) x (30/2)] / 210
= 90 / 210
= 9 /12