Question 21670: HI, I have a story problem I'm having a difficult time with, I was wondering if somebody could help me?
Question: A bag contains 6 red marbles, 9 blue marbles, and 5 green marbles. You withdraw one marble, replace it, and then withdraw another marble. What is the probability that you do NOT pick two green marbles?
thank you,
chris
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! A bag contains 6 red marbles, 9 blue marbles, and 5 green marbles. You withdraw one marble, replace it, and then withdraw another marble. What is the probability that you do NOT pick two green marbles?
SINCE THE MARBLES ARE REPLACED ,WE HAVE ALWAYS SAME NUMBER OF EACH COLOURED MARBLE IN THE BAG.THAT IS PROBABILITY OF DRAWING A PARTICULAR COLOUR MARBLE IS CONSTANT IN EACH PICK.LET US FIND THEM FIRST
TOTAL MARBLES = 6 RED+9 BLUE+5 GREEN =20
PROBABILITY OF DRAWING GREEN =PG=5/20=1/4
PROBABILITY OF DRAWING BLUE.. =PB=9/20
PROBABILITY OF DRAWING RED =PR=6/20=3/10
THEY ARE CONSTANT ALWAYS......
NOW IT IS ASKED THAT WE SHOULD NOT PICK 2 GREENS IN 2 PICKS...SO LET US LIST ALL POSSIBILITIES......
1ST.GREEN..2ND...BLUE...=PROBABILITY=(1/4)(9/20)
1ST.GREEN...2ND...RED...=PROBABILITY=(1/4)(3/10)
1ST.BLUE..2ND....GREEN...=PROBABILITY=(9/20)(1/4)
1ST.BLUE...2ND....BLUE...=PROBABILITY=(9/20)(9/20)
1ST.BLUE...2ND....RED...=PROBABILITY=(9/20)(3/10)
1ST.RED...2ND...GREEN...=PROBABILITY=(3/10)(1/4)
1ST.RED...2ND...BLUE...=PROBABILITY=(3/10)(9/20)
1ST.RED...2ND...RED...=PROBABILITY=(3/10)(3/10)
HENCE FIND EACH OF THE ABOVE AND ADD THEM UP TO GET TOTAL PROBABILITY OF NOT PICKING 2 GREEN MARBLES...ANOTHER EASIER WAY WOULD BE TO FIND PROBABILITY OF DRAWING 2 GREENS AND SUBTRACTING THAT FROM TOTAL PROBABILITY OF 1 TO GET THE PROBABILITY OF NOT DRAWING 2 GREENS...
THOUGH THIS IS EASIER AND BETTER I DID THE OTHER LONGER WAY TO SHOW YOU THE LOGIC AND METHOD..SO THE ANSWER YOU GET IS
1ST.GREEN..2ND...GREEN...=PROBABILITY=(1/4)(1/4)=1/16
HENCE PROBABILITY OF NOT GETTING 2 GREENS IS =1-1/16=15/16
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