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?
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
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.
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
To find the probability of getting exactly 2 heads when tossing 4 coins, we can use the concept of combinations.
First, let's consider the total number of possible outcomes when tossing 4 coins. Each coin can have 2 outcomes (heads or tails), so the total number of outcomes for 4 coins is 2^4 = 16.
Next, let's determine the number of ways we can get exactly 2 heads. We can use combinations to calculate this. The formula for combinations is:
C(n, r) = n! / (r! * (n-r)!)
where "n" is the total number of items, and "r" is the number of items being selected.
In this case, we have 4 coins, and we want to select 2 heads. So the number of combinations of getting exactly 2 heads is:
C(4, 2) = 4! / (2! * (4-2)!) = 6
Finally, we can calculate the probability of getting exactly 2 heads by dividing the number of favorable outcomes (getting exactly 2 heads) by the total number of possible outcomes:
Probability of getting exactly 2 heads = Number of combinations / Total number of outcomes = 6 / 16 = 3 / 8 = 0.375 or 37.5%
Now, let's move on to the second question.
To find the probability of getting at least 2 heads when tossing 4 coins, we need to consider all the favorable outcomes where we get 2, 3, or 4 heads.
First, let's calculate the number of combinations for getting exactly 3 heads. Using the same formula as before:
C(4, 3) = 4! / (3! * (4-3)!) = 4
Next, let's calculate the number of combinations for getting exactly 4 heads:
C(4, 4) = 4! / (4! * (4-4)!) = 1
Now, we can calculate the total number of favorable outcomes (getting at least 2 heads) by adding up the number of combinations for getting exactly 2, 3, or 4 heads:
Total number of favorable outcomes = Number of combinations for 2 heads + Number of combinations for 3 heads + Number of combinations for 4 heads = 6 + 4 + 1 = 11
Finally, we can calculate the probability of getting at least 2 heads by dividing the number of favorable outcomes by the total number of possible outcomes (which is 16, as calculated before):
Probability of getting at least 2 heads = Number of favorable outcomes / Total number of outcomes = 11 / 16 = 0.6875 or 68.75%