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 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
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 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%
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
? 56Pr + 6 : 54Pr + 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
Required number of ways = 10C3
= 10!/ {3! (10 - 3)!} = 10! / (3! 7!) = (10 x 9 x 8) / (3 x 2) = 120
Required number of ways = 25C3
= (25 x 24 x 23) / (1 x 2 x 3) = 2300
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