# simple probability question

• Sep 19th 2009, 12:25 PM
rish
simple probability question
If I have two coins and I have the ability to land heads 60% of the time

What is the probability that I will land one heads and then one tails (40%)

probability of landing two heads would be 30% and two tails would be 20%, but what about one heads and one tails?

Thanks
• Sep 19th 2009, 12:40 PM
e^(i*pi)
Quote:

Originally Posted by rish
If I have two coins and I have the ability to land heads 60% of the time

What is the probability that I will land one heads and then one tails (40%)

probability of landing two heads would be 30% and two tails would be 20%, but what about one heads and one tails?

Thanks

$0.6 \times 0.4 = 0.24 = 24\%$

It's the chance of getting heads multiplied by the chance of getting tails. It's clear if you draw a tree diagram
• Sep 19th 2009, 12:50 PM
rish
but if it were a 50/50 chance then wouldn't the probability be 50% of landing on of heads and one of tails?
• Sep 19th 2009, 12:53 PM
Plato
Quote:

Originally Posted by rish
If I have two coins and I have the ability to land heads 60% of the time
What is the probability that I will land one heads and then one tails (40%)
probability of landing two heads would be 30% and two tails would be 20%, but what about one heads and one tails?

Will you please explain the question?
What does "two coins and I have the ability to land heads 60% of the time" mean?
• Sep 19th 2009, 12:54 PM
e^(i*pi)
Quote:

Originally Posted by rish
but if it were a 50/50 chance then wouldn't the probability be 50% of landing on of heads and one of tails?

If they're fair coins how does the following statement from the question come into it?

Quote:

If I have two coins and I have the ability to land heads 60% of the time
If you have two fair coins there is a 50% chance of the first one landing heads and a 50% chance it is tails.

If it tails we ignore it but if it's heads we toss the second coin which is the same as the first so there is a 50% chance of getting tails.

Therefore we multiply the first 50% by the second 50% to get 25%.

I should note this is for heads then tails.
• Sep 19th 2009, 01:03 PM
rish
Ok, lets make things easier by excluding the fact that there are two coins, lets just make it one.

In one case the coin lands on heads 50% of the time
For heads to land twice the probability would be 25%
For tails to land twice it would also be 25%
But for one of each the probability would be 50%

Now lets change the situation to make heads land 60% of the time
probability of landing heads twice would be 30%
probability of landing tails twice would be 20%
Now, what is the probability of landing one heads and one tails?
• Sep 19th 2009, 01:08 PM
Plato
Quote:

Originally Posted by rish
Now lets change the situation to make heads land 60% of the time
probability of landing heads twice would be 30%
probability of landing tails twice would be 20%
Now, what is the probability of landing one heads and one tails?

NO, probability of landing heads twice would be $(.6)^2=.36$.

AND probability of landing tails twice would be $(.4)^2=.16$
• Sep 19th 2009, 01:15 PM
rish
oh ok, you have to square them.. so then how would i calculate the probability of landing one of each?
• Sep 19th 2009, 01:22 PM
Plato
Quote:

Originally Posted by rish
oh ok, you have to square them.. so then how would i calculate the probability of landing one of each?

Because the event can happen in two ways, HT or TH, it is
$(.6)(.4)+(.4)(.6)=2(.24)=.48$

BTW: NOTE $.36+.48+.16=1$