Probablity Question Help (1 Viewer)

[MATHmaster]

Q10. a) There are 5 gold coins and 3 silver coins in a purse. Christine selects 2 of
these coins at random. Find the probability that:
i. She selects 2 gold coins.
ii. She selects 1 gold coins and 1 silver coin.
iii. She selects 2 gold coins if it is known that she selected at least I
gold coin.

Having trouble with (iii)
Any ideas?

 

[Beaconite]

Q10. a) There are 5 gold coins and 3 silver coins in a purse. Christine selects 2 of
these coins at random. Find the probability that:
i. She selects 2 gold coins.
ii. She selects 1 gold coins and 1 silver coin.
iii. She selects 2 gold coins if it is known that she selected at least I
gold coin.

Having trouble with (iii)
Any ideas?

is that a 1?

 

[MATHmaster]

Yep soz, typo

 

[Timske]

Yep soz, typo

(5/8)(4/7)(3/6) i think

 

[MATHmaster]

I'm having trouble with the 'at least one part'. This is what I got:

(5/8)(4/7)(47/56) = 0.3

I got 47/56 by making 1 - 3/8 x 3/7 = 47/56 the probability of her selecting at least one gold coin, but I don't know how to apply it in the question correctly :S

 

[MATHmaster]

Can anyone answer this??? pls

 

[jamesischool]

Q10. a) There are 5 gold coins and 3 silver coins in a purse. Christine selects 2 of
these coins at random. Find the probability that:
i. She selects 2 gold coins.
ii. She selects 1 gold coins and 1 silver coin.
iii. She selects 2 gold coins if it is known that she selected at least I
gold coin.

Having trouble with (iii)
Any ideas?

i. (5/8) * (5/8) = 25/40
ii. (5/8) * (3/8) = 15/40
iii. if she selected at least 1 gold coin already that means there are 4 gold and 3 silver left. so that gives her a 4/7 chance right?

not sure if my answers are correct but that's what i'd do

 

[Shadowdude]

Q10. a) There are 5 gold coins and 3 silver coins in a purse. Christine selects 2 of
these coins at random. Find the probability that:
i. She selects 2 gold coins.
ii. She selects 1 gold coins and 1 silver coin.
iii. She selects 2 gold coins if it is known that she selected at least I
gold coin.

Having trouble with (iii)
Any ideas?

iii.

We have selected one gold coin already. So there are 7 coins left in the purse.

Total number of possibilities for the second selection: 7.
Total number of 'successes' for the second selection, i.e. selects another gold coin: 4 (there are 4 gold coins left in the purse)

Probability is 4/7.

 

[jamesischool]

iii.

We have selected one gold coin already. So there are 7 coins left in the purse.

Total number of possibilities for the second selection: 7.
Total number of 'successes' for the second selection, i.e. selects another gold coin: 4 (there are 4 gold coins left in the purse)

Probability is 4/7.

^ woot i was right. iii is pretty much just a question with an easy answer and a question you need to read carefully

 

[MATHmaster]

i. (5/8) * (5/8) = 25/40
ii. (5/8) * (3/8) = 15/40
iii. if she selected at least 1 gold coin already that means there are 4 gold and 3 silver left. so that gives her a 4/7 chance right?

not sure if my answers are correct but that's what i'd do

how is P(2 gold coins) = 5/8 x 5/8, isn't it 5/8 x 4/7?
and how is P(1 gold and 1 silver) = 5/8 x 3/8, isn't it 5/8 x 3/7 + 3/8 x 5/7?

Thanks for the final part though!!
How

 

[jamesischool]

oh yea my bad i forgot about 1 coin being taken out for both of them, and also forgot about the gold-silver and silver-gold
you're right for both of them

how is P(2 gold coins) = 5/8 x 5/8, isn't it 5/8 x 4/7?
and how is P(1 gold and 1 silver) = 5/8 x 3/8, isn't it 5/8 x 3/7 + 3/8 x 5/7?

Thanks for the final part though!!
How

 

[Indoz]

As you can see, I think the answers are i) 5/14, ii) 15/28, iii) 16/49

 

[Indoz]

View attachment 26262 As you can see, I think the answers are i) 5/14, ii) 15/28, iii) 16/49

sorry about the messy handwriting

 

[asianese]

I'm confused to the many solutions so here is my take

 

[barbernator]

Would the answer to iii) not be P(2 gold coins) / P(1 or 2 gold coins)

so (5/8)(4/7) / [(5/8)(4/7) + (3/8)(5/7) + (5/8)(3/7)]

 

[Carrotsticks]

We have G G G G G S S S

We want to find the probability of picking 2 gold coins, provided that we must already have at least 1.

If we are already GUARANTEED at least 1 gold coin, that means our next one MUST be a gold one too.

So in our bag, we have 4 golds remaining, and 3 silvers (total of 7).

We want to pick a gold, so 4/7.

 

[barbernator]

We have G G G G G S S S

We want to find the probability of picking 2 gold coins, provided that we must already have at least 1.

If we are already GUARANTEED at least 1 gold coin, that means our next one MUST be a gold one too.

So in our bag, we have 4 golds remaining, and 3 silvers (total of 7).

We want to pick a gold, so 4/7.

What I interpretted from the question is that the coins were chosen, and then is was said that there was at least 1 gold chosen. My question is, if we assume that the gold was chosen first then does that not change the odds?

When I looked at it, I just thought to take all the ways in which at least 1 gold can be chosen and that becomes our sample set, and from that take the ways in which we can choose 2 gold. hence

5C2 / [5C1x3C1 + 5C3] = 2/5 which is the same as (5/8)(4/7) / [(5/8)(4/7) + (3/8)(5/7) + (5/8)(3/7)] from tree diagram

 

[Carrotsticks]

We can't do 5C2 and that stuff, because the gold coins are (I assume) identical.

If they were NOT identical, then you would be correct.

 

[barbernator]

We can't do 5C2 and that stuff, because the gold coins are (I assume) identical.

If they were NOT identical, then you would be correct.

ahk alright.

why does this not work then?

P(2 gold coins) / P(1 or 2 gold coins)

 

[Aesytic]

i don't think the answer is purely 4/7, because the question says if at least one gold coin is selected. this doesn't mean that the gold coin was selected first; it could have been selected second, and all we know is that at least one was chosen, so i reckon barbenator's answer is correct