A problem.. there are 9 balls...one out of which is different..(lighter or heavier) have to find out in 3 measurements..which one is tht....
plzz help
1)make two packs of 4 balls. 1 ball will be left. if the two packs are equal, the the left one is the heavier ball. if the packs are unequal, then take the heavy pack 2) divide the pack of 4 balls into two packs of 2 balls each and measure. the pack which is heavier keep it aside. 3)take the heavy pack of 2 balls and weigh each ball. you will be able to find the heavier one
umm...it doesnt make that big a difference but it doesnt hurt to do some words either!!! u just need to comprehend the sentence carefully and u will get a fair idea of the so-called-complex-word....whilst in essays, u dont need to use flowery language or hi-fundu words in my opinion..your thought process must be clear and smooth..i got a 5.5 without using a single fancy word...
"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
"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
"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
total sample space = 2^6=64 probability that tails will be result atleast twice but not more than 5 times : 2 tails can occur in 6 flips in 6c2 = 15 ways similarly, 3 tails can occur in 6c3 =20 ways 4 tails can occur in 6c4 = 15 ways 5 tails can occur in 6c5 = 6 ways hence, total no. of ways = 15+20+15+6 = 56 ways hence, probability = 56/64 = 7/8 answer is C...
"THe members of the newest recruiting class of a certain military organization are taking their physical conditioning test, and those who score in the bottom 16 percent will have to retest. If the scores are normally distributed and have an arithmetic mean of 72, what is the score at or below which the recruits will have to retest? (1) There are 500 recruits in the class. (2) 10 recruits scored 82 or higher.
A certain law firm consists of 4 senior partners & 6 junior partners. How many different partners of 3 partners can be formed in which at least one member of the group is a senior partner (two groups are considered different if at least one group member is different.
Options 48 100 120 288 600
OA 100(Highlight sentence with mouse to see the answer)
Please give detailed explanation since my answer does not even figure in the options
A certain law firm consists of 4 senior partners & 6 junior partners. How many different partners of 3 partners can be formed in which at least one member of the group is a senior partner (two groups are considered different if at least one group member is different.
Options 48 100 120 288 600
OA 100(Highlight sentence with mouse to see the answer)
Please give detailed explanation since my answer does not even figure in the options
hey rohit, here's my solution: we have 4S and 6J, lets say. total number of groups would be 10c3 = 120. we need those that have atleast one S, so count out the ones with no S, that is 6c3 = 20.
the answer would be 120 - 20 = 100. let me know if you need a better explanation.
A certain law firm consists of 4 senior partners & 6 junior partners. How many different partners of 3 partners can be formed in which at least one member of the group is a senior partner (two groups are considered different if at least one group member is different.
Options 48 100 120 288 600
OA 100(Highlight sentence with mouse to see the answer)
Please give detailed explanation since my answer does not even figure in the options
hmm...the group consists of 3 partners out of which atleast one member is a senior partner
case 1 : 1 senior, 2 junior
1 senior can be chosen out of 4 in 4 ways 2 juniors can be chosen out of 6 in 6c2 = 15 ways so total no. of ways = 60
case 2 : 2 senior, 1 junior
2 seniors can be chosen out of 4 in 4c2 = 6 ways 1 junior can be chosen out of 6 in 6 ways so total no. of ways = 36
"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
Another way to solve it is as follows The probability required is actually 1-(probability the tails comes 0 times or 1 time or 6 times) prob that it comes 0 times is 1/2^6 or 1/64 prob that it comes 1 time is 6/64 and prob that is comes 6 times is also 1/64
so the terms in the bracket is 1/64+6/64 +1/64=8/64 or 1/8 hence required ans is 1-(1/ = 7/8
(#697) I understand that the probability that it comes 0 times is the same that it comes 6 times 1/2 x 1/2 x 1/2 x 1/2 x1/2 x 1/2= 1/64, but could you pls. explain in detail the probability that it comes 1 time be 6/64?
(#697) I understand that the probability that it comes 0 times is the same that it comes 6 times 1/2 x 1/2 x 1/2 x 1/2 x1/2 x 1/2= 1/64, but could you pls. explain in detail the probability that it comes 1 time be 6/64?
look at it like this: there's 6c0 = 1 ways of choosing 0 tosses out of 6, so prob = 6c0 * (1/2)^6 = 1/64 there's 6c6 = 1 ways of choosing 0 tosses out of 6, so prob = 6c6 * (1/2)^6 = 1/64 there's 6c1 = 6 ways of choosing 0 tosses out of 6, so prob = 6c1 * (1/2)^6 = 6/64
(#697) I understand that the probability that it comes 0 times is the same that it comes 6 times 1/2 x 1/2 x 1/2 x 1/2 x1/2 x 1/2= 1/64, but could you pls. explain in detail the probability that it comes 1 time be 6/64?
hi,
yeah..it works the same way as you have already framed in ur previous post..the probability that it comes 0 or 6 times is 1/64= 1/2 x 1/2...6 times..this is because the verbal way of framing the same is tail / head comes in the first time and second time and third time....and sixth time. The operator 'and' stands for the multiplication operator 'x'. Similarly, the probability of tail coming once can be framed verbally as an event of tail coming the first time or second time or third time or...or sixth time. The operator 'or' stands for the addition operator '+'. Hence, the prob is 1/64+ 1/64 +1/64 +1/64 +1/64+1/64=6/64
Hope this explanation suffices. Else write back. Problems on probability can be solved much efficiently if we try to frame the problem statement or required event statement in logical sentences and replace operators with mathematical ones.
"On Monday a certain animal shelter housed 55 cats and dogs. By Friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday"? A) 11 B) 12 C) 13 D) 14 E) 20
= 7/8