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Probability of a Simple event A package contains 7 peanut

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Senior Manager
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Probability of a Simple event A package contains 7 peanut [#permalink] New post 18 Feb 2004, 10:00
Probability of a Simple event

A package contains 7 peanut cookies, 5 almond cookies, 2 chocolate-chip cookies, and 6 pecan cookies. Find the probability of selecting 3 cookies randomly in which 2 cookies are almond and 1 cookie is in a different flavor.

Probability of Independent Events


A bag contains 3 purple marbles and 7 pink marbles. Let each marble selected is put back into the bag before the next marble is selected. Find the probability of selecting 1 purple marble first and 1 pink marble next.

Probability of Dependent Events

A bag contains 3 purple marbles and 7 pink marbles. Let each marble selected is not put back into the bag before the next marble is selected. Find the probability of selecting 1 purple marble first and 1 pink marble next.

Probability of Mutually Exclusive Events


A box contains 3 yellow golf balls, 2 blue golf balls, and 6 white golf balls. Find the probability of selecting 5 golf balls in which at least 3 of them are white.
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 [#permalink] New post 03 Sep 2004, 11:25
Wanted to repost this as well to see what you guys thought.
For the first question I got
5/20*4/19*15/18=5/114.

Does that work for anyone else?
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 [#permalink] New post 03 Sep 2004, 11:33
Could someone please work through the final problem. I can calculate the number of total possibilities 11C5=463. However I am not sure how to calculate the numerator. 5C3, 5C4, 5C5 doesn't seem like the correct way because we have 6 white balls and I don't think it takes into account that there are more white than any other color.
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 [#permalink] New post 03 Sep 2004, 12:28
1. (5C2 * 15C1)/20C3 = 5/38

2. 3C1/10C1 * 7C1/10C1 = 3/10 * 7/10 = 21/100

3. 3/10 * 7/9 = 21/90

4. 6C3*5C2/11C5 + 6C4*5C1/11C5 + 6C5*5C0/11C5 = 281/462
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 [#permalink] New post 05 Sep 2004, 00:43
can someone explain the 1st and the 4th one ?
y is the answer for 1st not 5/114
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 [#permalink] New post 05 Sep 2004, 04:10
1. There are 5 almond cookies - so 2 of it can be choosen in 5C2 ways. There are remaining 15 other type cookies. 1 of it can be choosen in 15C1 ways. So three cookies - 2 of almond and 1 of the other type can be choosen in 5C2*15C1 ways. This the numerator.

To choose 3 cookies out of 20 cookies mean that the total no. of possibilities is 20C3 ways. This is the denominator.

hence (5C2*15C1)/20C3 = (10 * 15)/[(20*19*18)/6] = 5/38.

2. Atleasst 3 white of the 5 picked - means it can be 3 white, 4 white or all 5 white.

If we pick 3 white: then we have (applying the same fundamentals used in 1), (6C3*5C2)/11C5

If we pick 4 white: then we have: (6C4*5C1)/11C5

If we pick all 5 white: then we have: (6C5*5C0)/11C5

Since it can be either 3 or 4 or 5 whites - we add the above three. Hence 281/462.
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 [#permalink] New post 07 Sep 2004, 01:04
venksune wrote:
1. (5C2 * 15C1)/20C3 = 5/38

2. 3C1/10C1 * 7C1/10C1 = 3/10 * 7/10 = 21/100

3. 3/10 * 7/9 = 21/90

4. 6C3*5C2/11C5 + 6C4*5C1/11C5 + 6C5*5C0/11C5 = 281/462


for 4th -
P(picking atleast 3 white balls)
= 1-P(no white balls) = 1 - 5C5/11C5 = 1 - 1/462 = 461/462
what am I missing? Cant we interprete the statement to mean no white balls?
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 [#permalink] New post 07 Sep 2004, 01:38
Stuti,
No, we cant apply the complementary principle in this case. Let me explain...

Note that the complementary event E* of an event E is defined as the event E not happening. If you are not confused by the statement I will be surprised. Mathematically P(E*) = 1- P(E). Wow, sometimes math is more clear. Let us see a couple of examples where this principle is best applicable.

ex.1. A bag contains 6 red balls, 8 blue balls and 5 white balls. When two balls are drawn at random, what is the probability that the two balls are a) of the same colour and b) are of different colors.

Look at this question carefully. You can apply complementary event principle conveniently for b) if you have the answer for a).

read it this way...
P(Both balls belonging to different colors) = 1 - [P(both balls belonging to same color)]

As a addl. note, let me solve a) here. Total ways in which two balls can be drawn from the total of 19 balls = 19C2 ways. This is the denominator.

The numerator would be 6C2/19C2 + 8C2/19C2 + 5C2/19C2. Got it. Ok.

So b) will be 1 - [6C2/19C2 + 8C2/19C2 + 5C2/19C2]

Let me give one more example to use the complementary strategy.

ex 2: In a series of one days internations between India and Australia, the probability of India winning or drawing is say 1/3 and 1/4 respectively (in reality the probability of India winning can be 0, but let us presume that we can win or atleast draw). If we are asked to find what is the probability of India loosing to Australia - we can conveniently use the complementary approach. How? here it goes..

Prob of a loss = 1- (Prob of a win + Prob of a draw) = 1- (1/3+1/4) = 5/12.

Hope this was useful.
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 [#permalink] New post 07 Sep 2004, 03:17
venksune wrote:
Stuti,
No, we cant apply the complementary principle in this case. Let me explain...

Note that the complementary event E* of an event E is defined as the event E not happening. If you are not confused by the statement I will be surprised. Mathematically P(E*) = 1- P(E). Wow, sometimes math is more clear. Let us see a couple of examples where this principle is best applicable.

ex.1. A bag contains 6 red balls, 8 blue balls and 5 white balls. When two balls are drawn at random, what is the probability that the two balls are a) of the same colour and b) are of different colors.

Look at this question carefully. You can apply complementary event principle conveniently for b) if you have the answer for a).

read it this way...
P(Both balls belonging to different colors) = 1 - [P(both balls belonging to same color)]

As a addl. note, let me solve a) here. Total ways in which two balls can be drawn from the total of 19 balls = 19C2 ways. This is the denominator.

The numerator would be 6C2/19C2 + 8C2/19C2 + 5C2/19C2. Got it. Ok.

So b) will be 1 - [6C2/19C2 + 8C2/19C2 + 5C2/19C2]

Let me give one more example to use the complementary strategy.

ex 2: In a series of one days internations between India and Australia, the probability of India winning or drawing is say 1/3 and 1/4 respectively (in reality the probability of India winning can be 0, but let us presume that we can win or atleast draw). If we are asked to find what is the probability of India loosing to Australia - we can conveniently use the complementary approach. How? here it goes..

Prob of a loss = 1- (Prob of a win + Prob of a draw) = 1- (1/3+1/4) = 5/12.

Hope this was useful.


:good thanks!
  [#permalink] 07 Sep 2004, 03:17
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