The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target atleast three times, is

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

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

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

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

? ⁵⁶P_{r + 6} : ⁵⁴P_{r + 3} = 30800 : 1
? 56! / (50 - r)! = (30800 x 54!) / (51-r!)
? 56 x 55 = 30800/(51 - r)
? 51 - r = 10
? r = 41

Total no . of person = 8 + 7 = 15
No. of groups = ¹⁵C₆ = 15! / {6! (15 - 6)!} = 15! / (6! 9!)
= (15 x 14 x 13 x 12 x 11 x 10) / (6 x 5 x 4 x 3 x 2 x 1)
= 5005