The probability that a leap year selected at random contains 53 Sundays, is

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%

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.

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

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

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.

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

Required Probability
Given, n = 5 and r = 3
Then, Success P = 3/4
Failure, q = 1 - 3/4 = 1/4
Man hit the target thrice
= ⁵c₃ (3/4)³ (1/4)² + ⁵c₄ (3/4)³ (1/4) + ⁵c₃(3/4)⁵
= (270/1024) + (405/1024) + (243/1024)
= 918/1024
= 459/512

Total number of favourable outcomes n(S) = 6³ = 216
Combinations of outcomes for getting sum of 15 on uppermost face = (4, 5, 6), (5, 4, 6), ( 6, 5, 4), (5, 6, 4), (4, 6, 5), (6, 4, 5), (5, 5, 5),(6, 6, 3), (6, 3, 6), (3, 6, 6)
Now, outcomes on which first roll was a four, n(E) = (4, 5, 6),(4, 6, 5)
? P(E) = n(E)/n(S) = 2/216 =1/108

Number of 4- digit numbers which are formed with 1, 3, 5, 7, 9 = ⁵P₄
= 5 x 4 x 3 x 2 = 120 = n (s)

Number of 4 digit number which are formed with 1, 3, 5, 7, 9 and are divisible by 5 = ⁴P₃
= 4 x 3 x 2 = 24 = n (E)

? p(E) = n(E)/n( s) = 24/120
= 1/5

Total number of possible outcomes
= ¹²C₃ = 220
Number of events which do not contain blue marbles (3 marbles out of 7 marbles) = ⁷C₃ = 35
? Required probability = 1 - 35/220 = 37/44