Help!!!! 3 followup ques from arun sharma lod2.. Not getting the hint given..Kindly help!!!!!
ques-from 4 men nd 4 ladies a committee of 5 is to be formed. find the num of ways doing so if the committee consisits of president,a vice-president and three secretaries?
a-8p5
b-1120
c-4c2 x 4c3
d-720
(ii) in above ques,what wil be the num of ways of selecting the committee with atleast 3 women such that atleast one women hold the post of either a president or a vice-president..?
a-420
b-610
c-256
d-512
(iii) find da num of ways of selecting the committee with a max of 2 women and having at the max one women holding one of the two posts on da committee.
a-16
b-512
c-608
d-256
e-324
Help!!!! 3 followup ques from arun sharma lod2.. Not getting the hint given..Kindly help!!!!!
ques-from 4 men nd 4 ladies a committee of 5 is to be formed. find the num of ways doing so if the committee consisits of president,a vice-president and three secretaries?
a-8p5
b-1120
c-4c2 x 4c3
d-720
(ii) in above ques,what wil be the num of ways of selecting the committee with atleast 3 women such that atleast one women hold the post of either a president or a vice-president..?
a-420
b-610
c-256
d-512
(iii) find da num of ways of selecting the committee with a max of 2 women and having at the max one women holding one of the two posts on da committee.
a-16
b-512
c-608
d-256
e-324
My takes are 1120, 512 and 512
(1) Precident can be selected in 8 ways, vice-precident can be selected in 7 ways and remaining 3 secretaries can be selected in 6C3 ways.
Total 8*7*6C3 ways = 1120 ways
(2) We can have 4 women or 3 women.
Case 1: 4 women.
Women can be selected in 1 way and one man can be selcted in 4 ways. Total 4 ways.
Now, precident can be selcted in 5 ways and vice-precident can be selcted in 4 ways. As we have only one man, women will have at least one of the two posts.
So, total 4*5*4 ways = 80 ways
Case 2: 3 women
3 women can be selcted in 4C3 ways and 2 men can be selcted in 4C2 ways
Total 4C3*4C2 ways = 24 ways
Precident can be selected in 5 ways and vice-precident can be selcted in 4 ways. But we need to remove cases when both both posts go to men.
Such cases will be 2 as either man can be the precident.
So, total ways = 24*(5*4 - 2) = 432
So, in all we have 80 + 432 = 512 ways
(3) Question 3 is sam eas question 2.
We need at most 2 women => At least 3 men
Women holding at most one post => Men must hold at least one post
So, again we get 512 ways
Pls Explain:
1) From a bag containing 4 white and 5 black balls a man draws 3 at random. what are the odds against these being selected? (pls explain what they mean by odds against these being selected)
a) 5/37 b) 37/5 c)11/13 d)13/37
2) In a horse race there were 18 horses numbered 1-18. The probability that horse 1 would win is 1/6, that 2 would win is 1/10 and that 3 would win is 1/8. Assuming that tie is not possible, find the chance that one of the three will win.
a)47/120 b)119/120 c)11/129 d)15/216
3) A husband and wife appear in an interview for two vacancies for the same post. The probability of husband's selection is 1/7 and of wife is 1/5. what is the probability that both of them will be selected?
a)1/35 b)2/35 c)3/35 d)1/7
4)Amit throws three dice in a special game of ludo. If it is known that he needs 15 or higher in this throw to win then find the chance of his winning the game
a)5/54 b)17/216 c)13/216 d)15/216
Explanation pls how to go about these question:
1) In how many ways can a person send invitation cards to 6 of his friends if he has four servants to distribute the cards?
a)6^4 b)4^6 c)24 d)none of these.
2) In how many ways can a prize be distributed to 8 students if each can get any number of prizes?
a)5^8 b)8^5 c)40 d)none of these.
Explanation pls how to go about these question:
1) In how many ways can a person send invitation cards to 6 of his friends if he has four servants to distribute the cards?
a)6^4 b)4^6 c)24 d)none of these.
2) In how many ways can a prize be distributed to 8 students if each can get any number of prizes?
a)5^8 b)8^5 c)40 d)none of these.
1) For each of the 6 friends he has 4 options, so it will be 4^6 ways.
It won't be 6^4, as it means that for every servant we have 6 options (which means its also possible that all the servants are going to the same friend which doesn't make any sense)
2) I think you forgot to mention the number of prizes (5 in this case)
So, each prize can go to any of the 8 students. That means 8^5 ways.
Here also it can not be 5^8, as then it means that every student can get any of the prize (which includes cases when one prize is given to more than one student which shouldn't be the case)
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short-cut to find 40!
Pls Explain:
1) From a bag containing 4 white and 5 black balls a man draws 3 at random. what are the odds against these being selected? (pls explain what they mean by odds against these being selected)
a) 5/37 b) 37/5 c)11/13 d)13/37
I think this is what are the odds against BLACK being selected
=> Probability = 5c3/9c3 = 5*4*3/9*8*7 = 5/42
Now, 'Odds FOR' means favorable case => favorable/(total-favorable)
And 'Odds AGAINST' means, unfavorable case => (total-favorable)/favorable
=> Favourable = 5, Total = 42
=> Odds Against = (42 - 5)/5 = 37/5 -> Option B
2) In a horse race there were 18 horses numbered 1-18. The probability that horse 1 would win is 1/6, that 2 would win is 1/10 and that 3 would win is 1/8. Assuming that tie is not possible, find the chance that one of the three will win.
a)47/120 b)119/120 c)11/129 d)15/216
I think here the outcomes are not dependent.
Hence, probability = 1/10 + 1/6 + 1/8 = (12+20+15)/120 = 47/120
@Chill Bhai - Please yeh wala explain ki jiye naa. Maine 2-3 alag methods se try kiya lekin options main nahi tha jo mujhe aiya. Phir yeh wala hit and try se aiya hai. Iska logic please explain kijiye :)
3) A husband and wife appear in an interview for two vacancies for the same post. The probability of husband's selection is 1/7 and of wife is 1/5. what is the probability that both of them will be selected?
a)1/35 b)2/35 c)3/35 d)1/7
P(A) = Probability that both husband and wife are not rejected
P(B) = Probability that husband is selected and wife is rejected
P(C) = Probability that wife is selected and husband is selected
P(X) = probability of husband's selection = 1/7, P(X') = probability of husband's rejection = 6/7
P(Y) = probability of wife's selection = 1/5, P(Y') = probability of wife's rejection = 4/5
=> P(A) = 1 - P(X')*P(Y') = 1 - (6/7 * 4/5) = 1 - 24/35 = 11/35
=> P(B) = P(X)*P(Y') = 1/7 * 4/5 = 4/35
=> P(C) = P(Y)*P(X') = 1/5 * 6/7 = 6/35
=> Total Probability = P(A) - P(B) - P(C) = 11/35 - 4/35 - 6/35 = 1/35
4)Amit throws three dice in a special game of ludo. If it is known that he needs 15 or higher in this throw to win then find the chance of his winning the game
a)5/54 b)17/216 c)13/216 d)15/216
Total outcomes = 6^3 = 216
Favourable outcomes = sum of 15 or higher
=> 6 6 6 = 1 way
=> 6 6 5 = 3 ways
=> 6 6 4 = 3 ways
=> 6 6 3 = 3 ways
=> 6 5 5 = 3 ways
=> 6 5 4 = 6 ways
=> 5 5 5 = 1 way
=> total favourable outcomes = 1+3+3+3+3+6+1 = 20
=> total probability = 20/216 = 5/54
nitya2903 Saysshort-cut to find 40!
What do you want to know about 40! ? 😲 It'll be a very huge number. You'll not be asked in any entrance exam.
since tie is not possible, all the probabilities will be added...u r correct. I hope that soves ur dilemma!!!
Q :- there are 3 teachers and 6 students and a committee is to be chosen of 3 members in which 1 teacher must be included
a :- 9c3-6c3
but i want to ask that y answer is not 3c1.8c2
plzz tell me
thanks
Q :- there are 3 teachers and 6 students and a committee is to be chosen of 3 members in which 1 teacher must be included
a :- 9c3-6c3
but i want to ask that y answer is not 3c1.8c2
plzz tell me
thanks
It is like this:
Maximum cases = 9C3
We do not case of all three students i.e. 6C3
Hence, 9C3 - 6C3
But if we do; 3C1*8C2; there will be some overcounting.
Say we choose T1 in 3C1. And then (T2, T3) in 8C2 ...... T1, T2, T3 are 3 teachers
And in next round, we choose T2 in 3C1 and (T1, T3) in 8C2. So, we get repeat cases. Hence, some overcounting. So, we should do it like (9C3 - 6C3)
Q :- there are 3 teachers and 6 students and a committee is to be chosen of 3 members in which 1 teacher must be included
a :- 9c3-6c3
but i want to ask that y answer is not 3c1.8c2
plzz tell me
thanks
here atleast 1 teacher must be included. In 3c1*8c2 there are repetitions involved.
the other way can be 3c1*6c2 + 3c2*6c1 + 3c3 = 45+18+1 = 64 = 9c3-6c3
In a bag there are 3 green coloured balls, 5 red coloured balls, 2 blue coloured balls, 6 black coloured balls.
1)If 5 balls are selected at random from the bag,wht is the probability that the selection contains 2 blue balls and 3 red balls?
a)7C5/16C5 b)20/16C5 c)10/16C5 d)30/16C5 e)none
2)If 4 balls are selected at random from the bag then wht is the probability that the selection contains either red balls or black balls?
a)10C4/16C4 b)(5C4*6C4)/16C4 c)(5C4 +6C4)/16C4 d)11C4/16C4 e)none
i got ma ans as:
1) b
2) e
plzzzzzzzzzzzzz solve n c wht ans u people get!!!
-------------------repeated post------------------------
In a bag there are 3 green coloured balls, 5 red coloured balls, 2 blue coloured balls, 6 black coloured balls.
1)If 5 balls are selected at random from the bag,wht is the probability that the selection contains 2 blue balls and 3 red balls?
a)7C5/16C5 b)20/16C5 c)10/16C5 d)30/16C5 e)none
3 green coloured balls,
5 red coloured balls,
2 blue coloured balls,
6 black coloured balls => total 16 balls
=> 2c2*5c3/16c5 = 10/16c5
2)If 4 balls are selected at random from the bag then wht is the probability that the selection contains either red balls or black balls?
a)10C4/16C4 b)(5C4*6C4)/16C4 c)(5C4 +6C4)/16C4 d)11C4/16C4 e)none
A small doubt in this sum.
If the question means that only red or only black balls are selected, then the prob = (5c4+6c4)/16c4
But if the question means that the selection contains red or black with combination of other balls, then: /16c4
either c or e
please answer dis-
how many numbers which are divisible by 5 can be formed using the digits 0 to 6 if no digits occur more than once in any number?
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