four cards are drawn at random (without Replacement) from a pack of 8 cards numbered 1 to 8?

Question

calculate the probability both 1 and 8 will be drawn

Answer 1

the number of ways 4 cards can be drawn = 8C4 = 8*7*6*5/(1*2*3*4)
if 1 and 8 both are there we have 2 other cards out of 6. Number of ways = 6C2 = 6*5/2

so probability = (6*5)/2/((8*7*6*50/24)
= 24/(2*8*7) = 12/56 = 3/14 or 300/14 % = 150/7 = 21.4 %

Answer 2

Probability problems can be a bit tricky and most people would be confused and then answer a different number of ways.

The key is that we should analyze what is asked and start from there.

**If we are looking for the probability that BOTH 1 AND 8 will be drawn:

There are two events that can get both 1 and 8:

(1) get 1 & then 8
1/8 x 1/7 = 1/56
or
(2) get 8 & then 1
1/8 x 1/7 = 1/56

-- BOTH events will happen so we add the two events:
1/56 + 1/56 = 2/56 or
= 1/28
or approximately 3.6% which is a very small probability bacause it will naturally be harder to get BOTH numbers

*Of course, if the problem is stated differently, then we would get a different answer, example:

--Probability that 1 OR 8 will be drawn:

- we should use here a "negative" or reverse solution to make it easier and find the probability that 1 OR 8 will NOT be drawn which is:

6/8 * 5/7 * 4/6 * 3/5 = 360/1680 or 3/14

- therefore, the probability that 1 OR 8 will be drawn will be:

1 - 3/14 = 11/14
or approximately 78.6% which is a high probability since it would be quite easy to get either a 1 OR 8.

-- You should really just start by understanding what is asked in the problem so that you won't get confused.

Hope that helps.

Answer 3

The probability of not drawing an 8 or 1 would be 6/8 *5/7 *4/6 * 3/5 = 360/1680 or 3/14.

Therefore the probabilty of drawing 1 and 8 would be 11/14.

Answer 4

obviously the total # of draw possibilties is 8*7*6*5

from this deduct the number where both 8 and 1 are not present
6*5*4*3

also deduct the number where 1 is present but 8 is not ( 1 could be drawn either 1 st or second or 3 rd or 4 th) so it's

(6 * 5 * 4) * 4

the same # occurs for when 1 is not but 8 is

so altogether we have
8*7*6*5-6*5*4*3-6*5*4*8=360 where you have both 1 & 8
since 8*7*6*5=1680 the probability is then
360/1680=.214285714285714

don't listen to that imho above me he just dont know

Answer 5

the probability of 1 or 8 being drawn at the first attempt is 2 out of 8 which is 2/8=1/4. next, the probability of the other coming out next is 1 out of 7, (because one card has been drawn). multipy that together and you get 1/28, which is pretty low....as for the other 2 cards, the probability of getting any other number is one, so the answer stays the same.

Or you might want to try going number by number,i.e. starting with the probability of 1 followed by 8 and vice versa. you'll get the same answer.

Answer 6

the probability of getting a 1 and then an 8

1/8 * 1/7 = 1/56

the probability of getting an 8 and then a 1

1/8 + 1/7 = 1/56

the probability of getting either combination

2/56 = 1/28

Answer 7

the respond is B, and here is why. to start up with, you have 50 tickets, and 25 of them ar over 25, so your odds of pulling a intense fee tag on the 1st pull are 25:50, or a million:2. Now on the 2d pull there are 40 9 tickets left and 24 intense ones (considering you already pulledf a intense one. Your astonishing on the 2d pull are 24:40 9, and mixing that with the opportunities of the 1st pull, your odds of pulling 2 immediately intense numbers is 24/ninety 8. The third pull has 40 8 tickets left and 23 of them are larger than 25, so your odds indexed under are 23:40 8, which whilst blended with the different 2 pulls, makes the whole odds 552:4704, which simplifies right down to 23/196.

Answer 8

there isnt a 1 but the 8 can be about 4/52

Answer 9

12/13/14/15/16/17/18/
23/24/25/26/27/28/
34/35/36/37/38/
45/46/47/48/
56/57/58/
67/68/
78
1 in 28 odds. :D
well I am doing it the hard way because ive never figured out a probability formulae on my own...cool! I hope someone has a much better answer than me. :D
anyone have actual probability formula?
ooops! you said 4 cards can be drawn
grrr....
am lost sorry someone tell! lol

Answer 10

on first card 2/8 or 1/4 chance to get either 1 or 8

on second card 1/4*1/7=1/28 to get both

1/28