OA is 176.
You might want to research on how to get it.
provide your expalination
provide your expalination
Sorry if Repost
A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. what is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer?
Prob that Harry NOT chosen for Prez - 9/10
Prob that Harry chosen for Sec - 9/10*1/9
Prob that Harry chosen for Treas. - 9/10*8/9 *1/8
Prob that Harry chosen for either Sec or Trea : 1/10+1/10 = 1/5
Tell me I din goof up in this one!!! 😞
OA is 176.
You might want to research on how to get it.
Hmm...176?
And you agree with that Harsh ?
Nisha,
If right angle triangle is 30-60-90 , you can get the angle as 30 right? :)
Imagine the diagram... you will get this.
Other angle is 150.
Answer is 30.
Simple expalination:
Area of triangle = 1/2*side1*side2* sin(angle between S1 & S2)
now area of gm = 2 Ar of triangle
=> 6*6*Sin@ =18
=>Sin@=1/2
=>@=30 degrees.
Hmm...176?
And you agree with that Harsh ?
Nopes...thats why i posted it here...
pj02 SaysNopes...thats why i posted it here...
What is the source of this Question ..as well as the ans that u r telling.
rohitforever SaysWhat is the source of this Question ..as well as the ans that u r telling.
My friend gave me her notes ( notebook with formulae written in it )
in that book she had noted this problem with answer.
It could be wrong too :)
I just put it here because i found it nice and thought it will be good to share with you guys... ( Also i tht i am missing something so i am getting a diff answer , i tht someone will get 176 and will prove me wrong...
Anyways... Thanks for discussing...
Going to bed now.. c u all tomm.. Good Night!!!
Prob that Harry NOT chosen for Prez - 9/10
Prob that Harry chosen for Sec - 9/10*1/9
Prob that Harry chosen for Treas. - 9/10*8/9 *1/8
Prob that Harry chosen for either Sec or Trea : 1/10+1/10 = 1/5
Tell me I din goof up in this one!!! :-(
E is correct.
one more for you guys... This one is simple...
A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?
A) 5/8
B) 3/4
C) 7/8
D) 57/64
E) 15/16
Damn!!! I just took the angle to be 60 instead!!!
quant isnt made for me!! :banghead:
I believe you are only making silly mistakes, nothing else.
You have solved the previous questions(where you forgot to ad the velocity) and this question also correctly, but somehow in a jiffy you opt for wrong answer.
Only the belief make us strong or weak
I believe you are only making silly mistakes, nothing else.
You have solved the previous questions(where you forgot to ad the velocity) and this question also correctly, but somehow in a jiffy you opt for wrong answer.
Only the belief make us strong or weak :cheers:
Yeah.. I need to stop being careless & get my act together!!
Thanks!
one more for you guys... This one is simple...
A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?
A) 5/8
B) 3/4
C) 7/8
D) 57/64
E) 15/16
My ans is marked in bold..
bhai... ye permutation & combination badi gandi cheej hai
I solved the question like this...
p(e)=required probability=1-{prob of getting one Tail+prob of getting all six tails}
p(e)=1-{(6C5)/2^6 + (6C0)/2^6}
=1-7/64
=57/64:splat:
pls let me know of my mistakes,if any.
My friend gave me her notes ( notebook with formulae written in it )
in that book she had noted this problem with answer.
It could be wrong too :)
I just put it here because i found it nice and thought it will be good to share with you guys... ( Also i tht i am missing something so i am getting a diff answer , i tht someone will get 176 and will prove me wrong...
Anyways... Thanks for discussing...
Going to bed now.. c u all tomm.. Good Night!!!
I think i've come across this question somewhere..
but even I'm getting 470!!
wrong post
57/64 and 470% are my answers also..
My take is E
avg c for first 5 months= 64
total C target for next 2 months =150*2 +(150-64)*5 =730
Avr C target for 1 month =365.
% increase = (365-64)/64 *100 approx 470%
one more for you guys... This one is simple...
A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?
A) 5/8
B) 3/4
C) 7/8
D) 57/64
E) 15/16
My ans is marked in bold..
bhai... ye permutation & combination badi gandi cheej hai
I solved the question like this...
p(e)=required probability=1-{prob of getting one Tail+prob of getting all six tails}
p(e)=1-{(6C5)/2^6 + (6C0)/2^6}
=1-7/64
=57/64:splat:
pls let me know of my mistakes,if any.
I think i've come across this question somewhere..
but even I'm getting 470!! :shocked:
Puys solve this ..
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5 ?
A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3
Lock the posssibilities:-
Probability wen we have tails 0 times = 6C0 => 1
Probability wen we have tails 1 times = 6C1 => 6
Probability wen we have tails 2 times = 6C2
Probability wen we have tails 3 times = 6C3
Probability wen we have tails 4 times = 6C4 => 6C2
Probability wen we have tails 5 times = 6C5 => 6
Probability wen we have tails 6 times = 6C6 => 1
Prob. b/w 2 - 5 times = 6C2 + 6C3 + 6C4 + 6C5 / Adding all above prob.
= 56/64
= 7/8
So answer is C..
one more for you guys... This one is simple...
A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?
A) 5/8
B) 3/4
C) 7/8
D) 57/64
E) 15/16
Seems Simple one:-
5 possibilities:- 2,6 ; 3,5 ; 4,4 ; 5,3 ; 6,2
Ans = 2/5 (D)
Puys solve this ..
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5 ?
A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3
C. 7/8
Possibility if getting no tails = (1/2)^6 --I
Possibility if getting one tails = Possibility if getting 5 heads and one tails
= (1/2)^5 * (1/2). But this can appear in 6C1 ways because tails can take any of the 6 positions
So, Possibility if getting one tails = (1/2)^6 * 6C1 = 6/2^6 --II
Possibility if getting 6 tails = 1/2^6 --III
Adding I,II and III = (1/2)^6 + 6/2^6 + 1/2^6=8/2^6 = 1/8
Therefore , prossibility of getting 2,3,4, OR 5 tails = 1-1/8=7/8
one more for you guys... This one is simple...
A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?
A) 5/8
B) 3/4
C) 7/8
D) 57/64
E) 15/16
Please solve this.
Right triangle LMN is to be constructed in the xy-plane so that the right angle is at point L and LM is parallel to the x-axis. The x- and y- coordinates of L, M, and N are to be integers that satisfy the inequalities -3
(A) 72
(B) 576
(C) 4032
(D) 4608
(E) 6336