Probability and Ratio

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Probability and Ratio

by coolhabhi » Sat Mar 16, 2013 10:56 am
Six playing cards are lying face down on a table, two of them are kings. Two cards are drawn at random. Let 'a' denote the probability that at least one of the cards drawn is a king, and 'b' denote the probability of not drawing a king. The ratio a/b is
(A)≥ 0.25 and < 0.5
(B) ≥ 0.5 and < 0.75
(C) ≥ 0.75 and < 1.0
(D) ≥ 1.0 and < 1.25
(E) ≥ 1.25


Now I have done this:

For 'a':
probability that at least one of the cards drawn is a king = (2/6)(4/5) + (2/6)(1/5)
(2/6)(4/5) is for one king drawn and another card drawn
(2/6)(1/5) is for the case where both the cards drawn are kings

For 'b':
probability of not drawing a king = (4/6)(3/5)

So
a
- =
b

(2/6)(4/5) + (2/6)(1/5)
----------------------
(4/6)(3/5)

8 + 2
------
12

=>10/12 = 5/6.

Is my answer correct??

But the official answer is E
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by srcc25anu » Sat Mar 16, 2013 11:02 am
P(k,k)=2/6 * 1/6 = 1/18
p (~k, ~k) = 4/6 * 3/6 = 1/3 = b
Prob of atleast 1 king = 1 - 1/18 - 1/3 = 11/18 = a

a/b = 11/18 by 1/3 = 11/6 = ~1.8
Ans E

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by Anju@Gurome » Sat Mar 16, 2013 11:11 am
coolhabhi wrote:Six playing cards are lying face down on a table, two of them are kings. Two cards are drawn at random. Let 'a' denote the probability that at least one of the cards drawn is a king, and 'b' denote the probability of not drawing a king. The ratio a/b is
Number of ways to draw 2 cards out of 6 = 6C2 = 15

Number of ways to draw 2 kings out of 2 kings = 2C2 = 1
Number of ways to draw 1 king out of 2 kings and 1 card out of rest 4 = (2C1)*(4C1) = 2*4 = 8
Number of ways to draw no kings and 2 cards out of rest 4 = 4C2 = 6

Now, a = P(at least one king) = P(one king) + P(two kings) = (8 + 1)/15 = 3/5
And, b = P(no king) = 6/15 = 2/5

Hence, a/b = 3/2 = 1.5

The correct answer is E.
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by spark » Sat Mar 16, 2013 11:21 am
Since a is the probability of getting at least one king, there are three scenarios that provide the desired outcome: KO, OK, and KK, where K=king, O=other card.

You're missing the OK scenario, which adds (4/6)*(2/5) to a. The order matters when you look at all the possible outcomes in this way, so you have to consider OK separately from KO.

a = 3/5
b = 2/5

a/b = 3/2
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by Anju@Gurome » Sat Mar 16, 2013 11:27 am
coolhabhi wrote: For 'a':
probability that at least one of the cards drawn is a king = (2/6)(4/5) + (2/6)(1/5)
(2/6)(4/5) is for one king drawn and another card drawn
(2/6)(1/5) is for the case where both the cards drawn are kings

For 'b':
probability of not drawing a king = (4/6)(3/5)
As you are interpreting the situation as 1st draw then 2nd draw, you need to consider two scenarios here : (King in 1st draw and No king in 2nd draw) and (No king in 1st draw and King in 2nd draw)

You've only considered the first one.

Also, the things will be a lot more easier if you utilize the fact that, (a + b) will be equal to 1.
Now, b = (4/6)*(3/5) = 2/5.
Hence, a = 1 - 2/5 = 3/5
Hence, a/b = 3/2 = 1.5
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by coolhabhi » Sat Mar 16, 2013 12:07 pm
srcc25anu wrote:P(k,k)=2/6 * 1/6 = 1/18
p (~k, ~k) = 4/6 * 3/6 = 1/3 = b
Prob of atleast 1 king = 1 - 1/18 - 1/3 = 11/18 = a

a/b = 11/18 by 1/3 = 11/6 = ~1.8
Ans E
Have a look at this statement:
"Prob of atleast 1 king = 1 - 1/18 - 1/3 = 11/18 = a"
1 - 1/18 - 1/3 would be the probability of one king and not of "at least one king"

So Prob of atleast 1 king = 1 - p (~k, ~k)
Prob of atleast 1 king = 1 - 1/3 = 2/3

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by coolhabhi » Sat Mar 16, 2013 12:53 pm
Anju@Gurome wrote:
coolhabhi wrote:Six playing cards are lying face down on a table, two of them are kings. Two cards are drawn at random. Let 'a' denote the probability that at least one of the cards drawn is a king, and 'b' denote the probability of not drawing a king. The ratio a/b is
Number of ways to draw 2 cards out of 6 = 6C2 = 15

Number of ways to draw 2 kings out of 2 kings = 2C2 = 1
Number of ways to draw 1 king out of 2 kings and 1 card out of rest 4 = (2C1)*(4C1) = 2*4 = 8
Number of ways to draw no kings and 2 cards out of rest 4 = 4C2 = 6

Now, a = P(at least one king) = P(one king) + P(two kings) = (8 + 1)/15 = 3/5
And, b = P(no king) = 6/15 = 2/5

Hence, a/b = 3/2 = 1.5

The correct answer is E.
Now why did you take 6C2? I know you would say that the order matters. But please I am not strong on that part. Could you please explain it in length..

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by Anju@Gurome » Sun Mar 17, 2013 6:27 am
coolhabhi wrote:Now why did you take 6C2? I know you would say that the order matters. But please I am not strong on that part. Could you please explain it in length..
Because number of ways to draw to cards out of 6 is given by 6C2.

There is a basic difference between your approach and the one I posted.
While your approach is interpreting the situation as drawing first card then drawing the second one (hence order matters), my approach is interpreting the situation as simultaneous selection of two cards (hence order doesn't matter).

Let us take a simple example.
There are two cards and one of them is king. Now, What is the probability of drawing two cards at random so that one of them is king but the other is not?
As there are two cards and only one of them is king, whenever we'll draw two cards one of them will be king but the other will not be. Hence, the probability of such an event is 1.

Now follow your approach,
  • Required probability = (1/2)(1/1) = 1/2 ---> Wrong
Now follow my approach,
  • Total number of ways to select 2 cards out of 2 = 2C2 = 1
    Number of ways to select 1 king and 1 other card = (1C1)*(1C1) = 1*1 = 1
    Required probability = 1/1 = 1
Hope that helps.
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