There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random.
There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3.
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
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​ 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 . of three numbers, second is twice the first and is also thrice the third. if the average of the three numbers is 44, the largest number is:

option 1) 24                              option 2) 56option 3) 72               option 4) 108option 5)    36 given : five defective bulbs are accidently mixed with 20 good ones.to find : probability distribution from this lotformula used :five defective bulbs are accidently mixed with 20 good ones.total number of bulbs = 25x denote the number of defective bulbs drawnx can draw the value 0 , 1 , 2 , 3 , 4.since the number of bulbs drawn is 4, n = 4p(0) = p(getting a no defective bulb) = = = p(1) = p(getting 1 defective bulb and 3 good ones) = = p(1) = = p(2) = p(getting 2 defective bulbs and 2 good one) = p(2) = = = p(3) = p(getting 3 defective bulbs and 1 good one) = p(3) = = p(4) = p(getting all defective bulbs) = = = p(4) = the probability distribution table is as follows, we know that- - - - - (i)taking 3 as common, we get3; solving only bracket term first,here a = 2a, b = 3b; putting values in eq.i}applying }}ans: - 3}note: must memorize cubes upto 12 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 start with 50 easys. if you solve those quickly, then move into mediums. hards are pretty tricky tbh i just attempt then peek answers/memorize lol 1 a c d
2 a
3 c
4 b
5 c
i got a 100% * 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: $B$ if the value (or total amount) of these two kinds of coins is thirty dollars..
let n = # of nickles, so 500-n = # of dimes.
 
0.05n + 0.10(500-n) = 30.00 expressed the total value; solve for n and you'll then be able to give each count requested.