Answer to Question #25293 in Statistics and Probability for Ellie




 A club consists of four girls and six boys.
a. In how many ways can a committee of three people be chosen?
b. In how many ways can two boys and two girls be chosen to attend a competition?
c. In how many ways can three boys be chosen?
d. What is the probability that three boys will be chosen?
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you can pick 1 senior in 5 ways, 1 junior in 4 ways, 1 sophomore in 3 ways and i freshman in 2 ways.
in all you can make these selections i 5 * 4 * 3 * 2 ways = 120 ways.
 
please note: selecting a committee is a "combination" problem, not a "permutation". Total no. of possible outcomes = 30 {1, 2, 3, … 30}  

  

E ⟶ event of getting no. divisible by 3.  

  

No. of favourable outcomes = 10 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}  

  

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)  

  

P(E) = 10/30 = 1/3  

  

Bar E ⟶ event of getting no. not divisible by 3.  

  

P(Bar E) = 1 – P(E)  

  

= 1 - 1/3 = 2/3 given : let a , b and c be three students whose chances of solving a problem is given i.e , p(a) = , p(b) = and p(c) = . p() = , p() = and p()= to find: the probability that the problem is solved .here , p(the problem is solved) = 1 - p(the problem is not solved)= 1- p( )= 1 – [p( ) p( ) p( )]= 1- []= 1 - = therefore , the probability that the problem is solved is . total number of cards in a well-shuffled pack =52total number of queens and kings cards =4+4=8therefore, there are 44 cards (52−8) that are neither king nor queen.total number of favorable outcomes  =44  
probability is given by,p(a)=no. of total outcomesno. of possible outcomes​  
required probability =5244​=1311​ guys into blondes will vote blonde.guys into brunettes will vote brown There could be a variety of leaderboards. Like one of top points overall, and another for the most points gained in the previous 4 weeks or something. I may never top to overall leaderboard, but I have a chance of getting on a monthly leaderboard. and what about guessing? are you good at guessing? try to guess if i am or not, if you guess right you choose your reward ;p in order to have the election process more democratic there are few pre-requisites like:a. one vote, one value: to choose/elect the representative everyone must have one note with equal value. this is to avoid any type of conflicting situation.b. regular election: there should be regular elections in every democratic country so that people have their choice at regular intervals.c. free and fair election: in order to ensure a fair chance to the people there should be a free and fair election in every democratic country.d. political competition: in order to have democratic election parties, candidates should be free to contest elections and should offer some real choice to the voters. * If the number ends with a $0$ then there are $9$ choices for the first digit, $8$ for the second and $7$ for the third, which makes $1\times9\times8\times7= 504$ possibilities.
* If the number is even ending with something else than $0$ then there are $4$ choices for the last digit, $8$ choices for the first digit (no $0$ nor the last digit), $8$ for the second digit and $7$ for the third digit, which makes $4\times8\times8\times7 =1792$


Together, this gives $2296$ numbers with $4$ distinct digits that are even. Note that this does not allow leading $0$, as you see to want it based from the question.


Correct Answer: $B$ look into dr. shettles method, we were just talking on a post about this the other day. he tells you how to have a boy or girl.