Probability (1 Viewer)

[kazemagic]

Hai guys I got a question.
A hand of five cards contains a ten, jack, queen, king and ace. From the hand, two cards are drawn in succession, the first card not being replaced before the second card is drawn. Find the probability that the ace is selected.
I did:
1/5+1/4=9/20

but the answer said it was 2/5... it mentioned the first card isnt replaced b4 the second card is drawn so it shouldnt be 1/5+1/5 which equals to 2/5

Thanks la

 

[deswa1]

No you don't just add them- look at it like this:

1) You either get an ace on the first try which is 1/5
2) You don't get an ace on the first try which is 4/5 and then get an ace the second time which is 1/4. So the chance of getting the ace the second time around is 1/5 (4/5 x 1/4)

Add these together and you get 2/5.

The reason why your answer is wrong is because you say that the chance of getting the ace on the second 'round' is 1/4 but it isn't because this neglects the fact that you can draw it on the first -> you have to take that possibility into account

 

[enoilgam]

EDIT: Ignore I was wrong anyway.

 

[kazemagic]

No you don't just add them- look at it like this:

1) You either get an ace on the first try which is 1/5
2) You don't get an ace on the first try which is 4/5 and then get an ace the second time which is 1/4. So the chance of getting the ace the second time around is 1/5 (4/5 x 1/4)

Add these together and you get 2/5.

The reason why your answer is wrong is because you say that the chance of getting the ace on the second 'round' is 1/4 but it isn't because this neglects the fact that you can draw it on the first -> you have to take that possibility into account

Ohh i c thanks bro

 

[kazemagic]

Another question =.=
Two dice are thrown simultaneously. Find the probability of obtaining a three on the first throw.
So I did:
1) You can get a three on both dice which is 1/36
2) You can get three on the dice No.1 and a non-three on dice No.2 which is 5/36
3) You can get three on dice No.2 and a non-three on dice No.1 which is 5/36

You add them all up and it's 11/36 but why does the answer say its 1/6 D:
Wats wrong with my logic
EDIT: Nvm I know wats wrong now D:

 

[kazemagic]

Sigh heres another one:
Two dice are thrown simultaneously. Find the probability of obtaining a four on the 2nd throw.
So I did:
1) Probability of not getting a four on the first throw = 5/6
2) Probability of getting a 4 on a throw = 1/6

Times that together = 5/36. The answer says its 1/6 D: I dun understand

 

[Demento1]

Sigh heres another one:
Two dice are thrown simultaneously. Find the probability of obtaining a four on the 2nd throw.
So I did:
1) Probability of not getting a four on the first throw = 5/6
2) Probability of getting a 4 on a throw = 1/6

Times that together = 5/36. The answer says its 1/6 D: I dun understand

Is this question from a maths txtbook? If so, which one and page number plox.

 

[kenkap]

Hai guys I got a question.
A hand of five cards contains a ten, jack, queen, king and ace. From the hand, two cards are drawn in succession, the first card not being replaced before the second card is drawn. Find the probability that the ace is selected.
I did:
1/5+1/4=9/20

but the answer said it was 2/5... it mentioned the first card isnt replaced b4 the second card is drawn so it shouldnt be 1/5+1/5 which equals to 2/5

Thanks la

no. of total combs- 5C2

two cards are chosen, one of them is ace so the remaining card has 4 possibilities....Prob- 4/5C2= 2/5

 

[Demento1]

Sigh heres another one:
Two dice are thrown simultaneously. Find the probability of obtaining a four on the 2nd throw.
So I did:
1) Probability of not getting a four on the first throw = 5/6
2) Probability of getting a 4 on a throw = 1/6

Times that together = 5/36. The answer says its 1/6 D: I dun understand

By reading what you've typed, the 1st dice thrown won't actually make any difference - hence you exclude that result. Thus really, the only probability that matters is the 2nd throw, which is 1/6 chance for a 4.

 

[kazemagic]

Is this question from a maths txtbook? If so, which one and page number plox.

yr 12 camb 3u, page 396, Q17)b

 

[deswa1]

Sigh heres another one:
Two dice are thrown simultaneously. Find the probability of obtaining a four on the 2nd throw.
So I did:
1) Probability of not getting a four on the first throw = 5/6
2) Probability of getting a 4 on a throw = 1/6

Times that together = 5/36. The answer says its 1/6 D: I dun understand

You are looking at these in the wrong way. Its just asking what's the chance of getting a four on the second throw. What happens on the first throw, we don't give a shit about. So we throw the first one and ignore it. Now we throw the second one and the question asks what is the chance that we get a four on this. And its obviously 1/6

 

[kazemagic]

By reading what you've typed, the 1st dice thrown won't actually make any difference - hence you exclude that result. Thus really, the only probability that matters is the 2nd throw, which is 1/6 chance for a 4.

You are looking at these in the wrong way. Its just asking what's the chance of getting a four on the second throw. What happens on the first throw, we don't give a shit about. So we throw the first one and ignore it. Now we throw the second one and the question asks what is the chance that we get a four on this. And its obviously 1/6

O kk thx

 

[kazemagic]

Uhh another one lol:
Two dice are thrown simultaneously. Find the probability of obtaining at least one two.
What I did:
1) Probability of getting a double two = 1/36
2) Probability of getting a two and non two = 5/6(probability of not getting a two on the 1st dice) x 1/6(probability of getting a two on 2nd dice) = 5/36

Add them together 5/36+1/36 = 1/6
The answers say its 11/36. Can someone explain to me why pls?
Thanks.

 

[bleakarcher]

Uhh another one lol:
Two dice are thrown simultaneously. Find the probability of obtaining at least one two.
What I did:
1) Probability of getting a double two = 1/36
2) Probability of getting a two and non two = 5/6(probability of not getting a two on the 1st dice) x 1/6(probability of getting a two on 2nd dice) = 5/36

Add them together 5/36+1/36 = 1/6
The answers say its 11/36. Can someone explain to me why pls?
Thanks.

You have to count the probability of that twice since it can occur such that the first dice rolls a 2 and the second dice does not OR the second rolls a 2 and the first dice does not. So you have 2*(5/36)+(1/36)=11/36

 

[kazemagic]

You have to count the probability of that twice since it can occur such that the first dice rolls a 2 and the second dice does not OR the second rolls a 2 and the first dice does not. So you have 2*(5/36)+(1/36)=11/36

ohhh alrite thx!

 

[kazemagic]

Two dice are thrown simultaneously. Find the probability of obtaining neither a one nor a four appearing.
I did:
1. Probability of not getting a one = 5/6
2. Probability of not getting a four = 5/6

Times that together = 25/36. The answer says its 4/9 though. Man im so bad at this

 

[Demento1]

Two dice are thrown simultaneously. Find the probability of obtaining neither a one nor a four appearing.
I did:
1. Probability of not getting a one = 5/6
2. Probability of not getting a four = 5/6

Times that together = 25/36. The answer says its 4/9 though. Man im so bad at this

Think about it this way, kaze. Both the dices cannot attain a 1 OR 4. Therefore, you have to consider that the probability for one dice not attaining this outcome listed above is 4/6 chance. Remember there are two dices.

 

[kazemagic]

Think about it this way, kaze. Both the dices cannot attain a 1 OR 4. Therefore, you have to consider that the probability for one dice not attaining this outcome listed above is 4/6 chance. Remember there are two dices.

fuark im so dumb, thanks bro !

 

[kazemagic]

Fifty tagged fish were released into a dam known to contain fish. later a sample of thirty fish was netted from this dam, of which eight were found to be tagged. Estimate the total number of fish in the dam just prior to the sample of thirty being removed.

What I did:
1. I let x = Total contained fish b4 the 50 tagged fish were added in.
2. I assumed 8/30 was the probability of getting a tagged fish.

I get this equation, and I cant solve it because the "x" gets eliminated:
8/30= (50+[8x/30])/(50+x)
(50+[8x/30]) = Total number of tagged fish
(50+x) = Total number of fish


Answer is 187 or 188, and i dunno how to get it :/

 

[Sy123]

Fifty tagged fish were released into a dam known to contain fish. later a sample of thirty fish was netted from this dam, of which eight were found to be tagged. Estimate the total number of fish in the dam just prior to the sample of thirty being removed.

What I did:
1. I let x = Total contained fish b4 the 50 tagged fish were added in.
2. I assumed 8/30 was the probability of getting a tagged fish.

I get this equation, and I cant solve it because the "x" gets eliminated:
8/30= (50+[8x/30])/(50+x)
(50+[8x/30]) = Total number of tagged fish
(50+x) = Total number of fish


Answer is 187 or 188, and i dunno how to get it :/

Simple:

Of the 8 fish we get 30 are tagged, hence we can assume the ratio of tagged : non tagged is:



Let x equal the number of fish in the damn after the fish were added:

Ratio of the tagged: non tagged should be the same ->