Question

HELP ME PLEASE PLEASE PLEASE....how do I know what formula to use? joint or disjoint events? this is the problem and a picture of what i have...Compute the probability of randomly selecting a two or three from a deck of cards (52 cards)...By the way, thanks @... WHAT IS THE PROBABILITY OF SELECTING A 2 OR 3 OR 4 FROM THE DECK OF CARDS? (WHICH FORMULA SHOULD I USE?

Answer 1

i think : you have the formula for choosing 2 cards among 52 is 52C2 and choosing 3 cards among 52 is 52C3 the "or" means "+" them together. I mean 52C2+52C3 is the solution for part a. we can do the same with part b and c whose cases are adding more option

Answer 2

lET ME SHOW YOU WHAT I HAVE...

Answer 3

ok

Answer 4

give me a minute

Answer 5

ok so these are the 3 problems

Answer 6

I think your formula is not for the probability problem. it cannot be that. i give you idea: how many way to choose 2 cards among 52 cards? it cannot be 2/52 =26. you have many ways to choose it. for example: with card 1, combines with card 2 you have 1 case, combines with card3 you have 2nd case, combines with card4 you have 3rd case so on and so on... So the probability to choose 2cards among 52 is 52C2 = (52*51)/2 *1=1326. and then 52C3 = (52*51*50)/ 3*2*1=22100. Now put the "or" means + between them, you have 1326+22100=23426 that is the result for a). I don't know whether it matches with your book's answer or not. if it is so, you continue the same in part b, c. if it is not. I am sorry, I cannot help

Answer 7

no sorry the question was to select a card with a number 2 on ti or a card with a number 3 on it...like that kind of problem..but i get what you said :) thanks a lot anyway....

Answer 8

oh. sorry, I misunderstand the problem. ok, ignore mine, let me a minute

Answer 9

ok dnt worry...and go for it

Answer 10

since the probability to pick a "2" card is 1/13 and the same with a "3" card, so just add them together to get 2/13

Answer 11

yeah it would be from 52 cards in total ( total amount in the deck) so what i did was right?

Answer 12

yeeees

Answer 13

yeaiii ok thanks a lot!

Answer 14

you are welcome