p = .5
1-p = .5
binary coefs 1 4 6 4 1
prob 0 head 4 tails = 1*.5^0*.5^4 = .0625
prob 1 head 3 tails = 4*.5^1*.5^3 = .25
prob 2 head 2 tails = 6*.5*2*.5^2 = .5625
prob 3 head 1 tail = .25
prob 4 head 0 tail = .0625
You can take it from there I think
When tossing 4 coins,what is the chance of getting exactly 2 heads?
When tossing 4 coins, what is the chance of getting at least 2 heads?
3 answers
The probablity of any specific sequence of heads and tails is (1/2)^4, because the probability of getting a head at each throw is 1/2. So, the probability of getting exactly 2 heads is 1/16 times the number of ways of having two heads in a sequence of 4 throws. A particular sequence can be HHTT where "H" represents heads and "T" represents tails. The total number of distinct sequences is then 4!/(2!*2!). There are n! distinct sequences of n different objects. If the two H's and the two T's were different, you would have had 4! different sequences. But the 2! ways of permuting the two T's don't lead to a different sequences, and the same is true for the 2! ways of interchanging the two H's. That's why you need to divide the 4! by 2!^2.
There are thus 6 ways to get precisely 2 heads, the probability is thus 3/8.
To compute the probability of at least 2 heads, note the symmetry between heads and tails; the probability of at least 2 tails is the same. If we add up both probabilities we get twice the desired answer, but we are then also counting all the possible outcomes but we then double count the probability of getting precisely 2 heads.
The sum of the probability of all the possible outcomes obviously equals 1. Double counting the probability of 2 heads means adding 3/8 to this, this yileds 1 + 3/8 = 11/8. Dividing this by 2 yields 11/16.
There are thus 6 ways to get precisely 2 heads, the probability is thus 3/8.
To compute the probability of at least 2 heads, note the symmetry between heads and tails; the probability of at least 2 tails is the same. If we add up both probabilities we get twice the desired answer, but we are then also counting all the possible outcomes but we then double count the probability of getting precisely 2 heads.
The sum of the probability of all the possible outcomes obviously equals 1. Double counting the probability of 2 heads means adding 3/8 to this, this yileds 1 + 3/8 = 11/8. Dividing this by 2 yields 11/16.
prob 0 head 4 tails = 1*.5^0*.5^4 = .0625
prob 1 head 3 tails = 4*.5^1*.5^3 = .25
prob 2 head 2 tails = 6*.5*2*.5^2 = .375
prob 3 head 1 tail = .25
prob 4 head 0 tail = .0625
You can take it from there I think
prob 1 head 3 tails = 4*.5^1*.5^3 = .25
prob 2 head 2 tails = 6*.5*2*.5^2 = .375
prob 3 head 1 tail = .25
prob 4 head 0 tail = .0625
You can take it from there I think
1 answer