math probability word problem

[Electrons]

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A carton contains 12 ipods, 3 of which are defective. If 4 ipods are sold at random, find the probability that exactly 1 will be defective?

Answer: 28/55
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How do you solve this the normal way without using combination?

This is what I have so far, but I'm missing considering something else here...or completely wrong setup cause it doesn't equal 28/55 . lol

3/12 * 9/12 * 8/11 * 7/10 =

 

[Streetwolf]

A carton contains 12 ipods, 3 of which are defective. If 4 ipods are sold at random, find the probability that exactly 1 will be defective?

Answer: 28/55
-----------------------
How do you solve this the normal way without using combination?

This is what I have so far, but I'm missing considering something else here...or completely wrong setup cause it doesn't equal 28/55 . lol

3/12 * 9/12 * 8/11 * 7/10 =

The normal way uses a combination. Otherwise you'll have to add and multiply.

Maybe you're looking for this ridiculously long way?

(3/12 * 9/11 * 8/10 * 7/9) + (9/12 * 3/11 * 8/10 * 7/9) + (9/12 * 8/11 * 3/10 * 7/9) + (9/12 * 8/11 * 7/10 * 3/9) = 28/55

That takes into account (in order) the first one being defective, the second one being defective, the third one being defective, and the fourth one being defective, while the other 3 are not defective. For example when I put (3/12 * 9/11 * 8/10 * 7/9), the 3/12 means I'm selecting one of the 3 defective ipods out of the 12 total, then I'm selecting one of the 9 working ipods out of the 11 remaining, then I'm selecting one of the 8 working ipods out of the 10 remaining, then I'm selecting one of the 7 working ipods out of the 9 remaining. Alternatively I could have first picked out a working ipod, then a defective ipod, then two working ipods. That's why you need 4 terms being added together.

With a combination you first choose the 3 working ipods out of the 9 with (9 choose 3) = 84. Then you choose the 1 defective ipod out of the 3 with (3 choose 1) = 3. Multiply these together and get 252. Then find out how many ways you can select 4 ipods from 12 with (12 choose 4) = 495. Put 252 over 495 and get 28/55.

So much simpler.

 

[Electrons]

nevermind...i figured it out. that is one sick problem. it is much longer to do it w/o combination. the setup w/o combination is longer.

i think this is how it should be w/o combination method ..not sure though.

(3/12 * 9/11 * 8/10 * 7/9) + (9/12 * 3/11 * 8/10 * 7/9) + (9/12 * 8/11 * 3/10 * 7/9) + (9/12 * 8/11 * 7/10 * 3/9) = 28/55

 

[Electrons]

The normal way uses a combination. Otherwise you'll have to add and multiply.

Maybe you're looking for this ridiculously long way?

(3/12 * 9/11 * 8/10 * 7/9) + (9/12 * 3/11 * 8/10 * 7/9) + (9/12 * 8/11 * 3/10 * 7/9) + (9/12 * 8/11 * 7/10 * 3/9) = 28/55

That takes into account (in order) the first one being defective, the second one being defective, the third one being defective, and the fourth one being defective, while the other 3 are not defective. For example when I put (3/12 * 9/11 * 8/10 * 7/9), the 3/12 means I'm selecting one of the 3 defective ipods out of the 12 total, then I'm selecting one of the 9 working ipods out of the 11 remaining, then I'm selecting one of the 8 working ipods out of the 10 remaining, then I'm selecting one of the 7 working ipods out of the 9 remaining. Alternatively I could have first picked out a working ipod, then a defective ipod, then two working ipods. That's why you need 4 terms being added together.

With a combination you first choose the 3 working ipods out of the 9 with (9 choose 3) = 84. Then you choose the 1 defective ipod out of the 3 with (3 choose 1) = 3. Multiply these together and get 252. Then find out how many ways you can select 4 ipods from 12 with (12 choose 4) = 495. Put 252 over 495 and get 28/55.

So much simpler.

I was writing up what I was guessing how it should be setup but you wrote it before me.

woo hoo!!! THanks for explaining the logic to confirm the reasoning. I got it right though - after fatal logic mistake.

It would have probably cost 10 minutes going this long route method.

Yea you are right, combination would have been faster. I noticed earlier in some regular problem, combination probability took longer. When it has this "exact" I guess I have to go the combination route. Is this good assumption?

Is it multiply and add because of the MUTUAL EXCLUSIVE?

 

[Electrons]

Another problem using combination method.

A woman holds 2 of total of 20 raffle tickets. If there are 2 winning tickets, determine the probability that she has exactly one.

Answer: 18/95

My setup...but logic is probably wrong somewhere

(2C1 x 18C2) / 20C2 = [(2 x (18*17)/2)] / [(20*19)/2] = 153/95

 

[Electrons]

Another problem using combination method.

A woman holds 2 of total of 20 raffle tickets. If there are 2 winning tickets, determine the probability that she has exactly one.

Answer: 18/95

My setup...but logic is probably wrong somewhere

(2C1 x 18C2) / 20C2 = [(2 x (18*17)/2)] / [(20*19)/2] = 153/95

I found what my mistake is. Arghhhhh. It should be

2C1 x 18C1 / 20C2 = 18/95

 

[drgreen]

Another problem using combination method.

A woman holds 2 of total of 20 raffle tickets. If there are 2 winning tickets, determine the probability that she has exactly one.

Answer: 18/95

Explanation:

There are two possible options.

We will sum the probability of each case

Case A: Ticket 1=win, Ticket 2=loss
(odds of drawing a winner)(odds of drawing a loser)
=(2/20)(18/19)=9/95

Case B: Ticket 1=loss, Ticket 2=win
(Odds of drawing a loser)(odds of drawing winner)
=(18/20)(2/19)=9/95

9/95+9/95=18/95

 

[harrygt]

A carton contains 12 ipods, 3 of which are defective. If 4 ipods are sold at random, find the probability that exactly 1 will be defective?

Answer: 28/55
-----------------------
How do you solve this the normal way without using combination?

This is what I have so far, but I'm missing considering something else here...or completely wrong setup cause it doesn't equal 28/55 . lol

3/12 * 9/12 * 8/11 * 7/10 =

C(3,1) * C(9,3) / C(12,4) = 28/55 is the correct answer

 

[Electrons]

New one:



If 5 cards are dealt, what is the probability of getting 4 of a kind?


Answer: 13/54,145


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I got this so far but can't figure out what the rest is.


(?C4 * ?C1) / 52C5

 

[Streetwolf]

New one:



If 5 cards are dealt, what is the probability of getting 4 of a kind?


Answer: 13/54,145


---------
I got this so far but can't figure out what the rest is.


(?C4 * ?C1) / 52C5

You got 52C5 so thats good.

You NEED to reason each problem out. TALK it over. What does 4 of a kind mean? It means you have 4 of the same RANK, each with a different SUIT, with one extra card (can be anything).

You know you need to choose a rank (ace, 2, 3, ...). Do this with 13C1. Now you need to select the suits for this rank. There are 4 cards of this rank, each with one of four suits, so you use 4C4. Finally you need to choose the last card. Since it can be ANY of the remaining 48, you simply multiply by 48.

Answer:

(13C1 * 4C4 * 48) / (52C5) = 624/2598960 = 13/54145.

 

[chyuk84]

I was just curious, where did u get these probability questions??

 

[drgreen]

Another way of looking at it.

There are five cases that satisfy the desired outcome: 1st, 2nd, 3rd, 4th, and 5th card being dissimilar.

lets analyze one possibility

CASE FIVE: last card dissimilar

the first card can be anything (52/52)
for the second card to be the same (3/51)
for the third card to be the same (2/50)
for the fourth card to be the same (1/49)
for the fourth card to be different (48/48)

(52/52)(3/51)(2/50)(1/49)(48/48)

there are five different cases so multiply that by 5.

(52/52)(3/51)(2/50)(1/49)(48/48)(5)=ANSWER

 

[Electrons]

You got 52C5 so thats good.

You NEED to reason each problem out. TALK it over. What does 4 of a kind mean? It means you have 4 of the same RANK, each with a different SUIT, with one extra card (can be anything).

You know you need to choose a rank (ace, 2, 3, ...). Do this with 13C1. Now you need to select the suits for this rank. There are 4 cards of this rank, each with one of four suits, so you use 4C4. Finally you need to choose the last card. Since it can be ANY of the remaining 48, you simply multiply by 48.

Answer:

(13C1 * 4C4 * 48) / (52C5) = 624/2598960 = 13/54145.

Thanks, that was an interesting one. I think I get the big picture now.

I was just curious, where did u get these probability questions??

They are not exact questions. I reword and paraphrase them from various resources.

Another way of looking at it.

There are five cases that satisfy the desired outcome: 1st, 2nd, 3rd, 4th, and 5th card being dissimilar.

lets analyze one possibility

CASE FIVE: last card dissimilar

the first card can be anything (52/52)
for the second card to be the same (3/51)
for the third card to be the same (2/50)
for the fourth card to be the same (1/49)
for the fourth card to be different (48/48)

(52/52)(3/51)(2/50)(1/49)(48/48)

there are five different cases so multiply that by 5.

(52/52)(3/51)(2/50)(1/49)(48/48)(5)=ANSWER

I would have missed that 52/52 or 48/48 doing it this way. Thanks for showing it this route.

I was wondering why they do some via combination way and some in non-combination way. Now I know it's just a preference and convenience. I can see why they prefer one over the other. It's just the logic can get me mixed up. I get the big pic now. It'll take sometime to digest it.

 

[Electrons]

Another problem using combination method.

A woman holds 2 of total of 20 raffle tickets. If there are 2 winning tickets, determine the probability that she has exactly one.

Answer: 18/95

My setup...but logic is probably wrong somewhere

(2C1 x 18C2) / 20C2 = [(2 x (18*17)/2)] / [(20*19)/2] = 153/95

If this problem asked to find the probability that she has a) at least 1 winning ticket and b) of 2 winning tickets

The setup would be these right?

a) 1 - (19/20)(18/19) = 1/10

b) 2/20 (1/19) = 1 / 190

 

[Streetwolf]

If this problem asked to find the probability that she has a) at least 1 winning ticket and b) of 2 winning tickets

The setup would be these right?

a) 1 - (19/20)(18/19) = 1/10

b) 2/20 (1/19) = 1 / 190

(a) should be 1 - (18/20)(17/19)
(b) looks fine.

 

[Electrons]

(a) should be 1 - (18/20)(17/19)

arghghh i missed it. Thanks for fixing. It matches the concept.