A paper bag contains seven cards, numbered 1 - 7. If you were to pick four cards without looking, what is the probability that they would all be odd-numbered cards?
a. 4/7
b. 1/35
c. 16/49
d. 3/7
2007-01-13 03:24:31 · 9 answers · asked by Hardcore 3 in Science & Mathematics ➔ Mathematics
no they will not be put back
2007-01-13 03:32:44 · update #1
one question: are the cards that are picked without looking be put back into the bag or whatsoever before withdrawing the next cards? because it's really important..answers will be different
ok....the total possibilty of four cards with 7 numbers=7x6X5X4
odd numbers are=1,3,5,7
so the possibility that all 4 cards are odd number = 4X3X2X1
so the probality will be (4X3X2X1)/(7x6x5x4)
just do the math
2007-01-13 03:30:32 · answer #1 · answered by feng ning 2 · 1⤊ 0⤋
if you don't put the card back...
so when you withdraw the first card, there are 4 odd number out of 7 so the prob is 4/7
the second card, 3 odd number out of 6 cards (3/6)
the third card, 2 out of 5
and the last card is 1 out of 4
so, the porbability is 4/7*3/6*2/5*1/4= 1/35...
if you put it back then it will always be 4/7 and it will be (4/7)^4 = 256 / 2401 which is not in the choice.
so, the answer must be B 1/35
2007-01-13 11:36:41 · answer #2 · answered by Imoet 2 · 2⤊ 0⤋
There were two probabilities:
Case (a): Can put back into paper bag
odd number = 4 cards
total = 7 cards
pick 4 cards =(1st)4/7 x(2nd)4/7 x(3rd) 4/7 x(4th) 4/7 = 256/2401
Case (b): Cannot put back into paper bag
1st : odd = 4 , total = 7
2nd : odd = 3 , total = 6
3rd : odd = 2 , total = 5
4th : odd =1 . total = 4
pick 4 cards =(1st)4/7 x(2nd)3/6 x(3rd)2/5 x(4th)1/4 = 1/35
In this question, only one answer there, so it's CASE B.
2007-01-13 11:52:28 · answer #3 · answered by happynanny03 1 · 1⤊ 0⤋
If all four cards are odd-numbered, then the cards must be 1,3,5,7.
This is one possibility for choosing 4 cards out of the 7.
How many possibilities are there? well, according to a well-known formula, this is the binomial coefficient choose(7,4) which is equal to 7! / (4! * 3!) = 5040 / (6 * 24) = 35.
so there are 35 possibilities, and we are interested in a particular one, so our chances are 1/35 - answer b.
2007-01-13 11:36:23 · answer #4 · answered by freq_fraq 1 · 2⤊ 0⤋
the answer is b.
in the first draw, there are 4 odd from 7 possible, so the probability of getting odd is 4/7
in the second draw there are 3 odd from 6 possible ... 3/6
then it is 2/5, then last draw is 1/4
multiply all of these probabilities: 4/7*3/6*2/5*1/4 = 1/35
2007-01-13 11:38:37 · answer #5 · answered by Matt 3 · 2⤊ 0⤋
4/7 since you have a chance to pick either 1 or 3 or 5 or 7 which are odd-numbered cards.
2007-01-13 11:31:42 · answer #6 · answered by fernando_007 6 · 0⤊ 2⤋
definitely a.4/7
2007-01-13 11:34:33 · answer #7 · answered by nawwar aqilah 1 · 0⤊ 2⤋
a
because my lucky number is 7
i did not choose d because 3 is my unlucky number
2007-01-13 11:33:18 · answer #8 · answered by srinu710 4 · 0⤊ 2⤋
fifty percent chances
2007-01-13 11:29:10 · answer #9 · answered by david j 5 · 0⤊ 4⤋