Find the probability of tossing exactly 2 heads on 3 tosses of a fair coin.
HHH,HHT,HTH,THH,TTH,THT,HTT,TTT-- 8 ways
3 have 2 heads
prob(2 out of 3 H) = 3/8
To find the probability of tossing exactly 2 heads on 3 tosses of a fair coin, we can use the binomial probability formula.
The formula for the probability of getting exactly k successes in n independent Bernoulli trials (each with probability of success p) is:
P(k) = (nCk) * (p^k) * ((1-p)^(n-k))
In this case, n = 3 (3 tosses of a coin), and k = 2 (exactly 2 heads).
The probability of getting a head on a fair coin toss is p = 1/2.
Using the formula, we can calculate the probability as follows:
P(2) = (3C2) * ((1/2)^2) * ((1-(1/2))^(3-2))
To calculate (3C2), we use the combination formula:
(3C2) = 3! / (2!(3-2)!) = 3
Now plugging in the calculated values:
P(2) = 3 * ((1/2)^2) * ((1-(1/2))^1)
P(2) = 3 * (1/4) * (1/2)^1
P(2) = 3 * (1/4) * 1/2
P(2) = 3/8
Therefore, the probability of tossing exactly 2 heads on 3 tosses of a fair coin is 3/8.
To find the probability of tossing exactly 2 heads on 3 tosses of a fair coin, we can use the concept of binomial probability.
The binomial probability formula is given by:
P(x) = nCx * p^x * q^(n-x)
Where:
- P(x) is the probability of getting exactly x successes,
- n is the total number of trials,
- x is the number of successes,
- p is the probability of success on a single trial,
- q is the probability of failure on a single trial (1 - p), and
- nCx is the binomial coefficient, also known as "n choose x," which is the number of ways to choose x objects from a set of n objects.
In this case, we want to find the probability of getting exactly 2 heads, so:
- n = 3 (total number of tosses)
- x = 2 (number of heads)
- p = 0.5 (probability of getting a head on a single toss, as the coin is fair)
- q = 1 - p = 1 - 0.5 = 0.5 (probability of getting a tail on a single toss, as the coin is fair)
Now, let's substitute the values into the formula and calculate the probability:
P(2) = 3C2 * (0.5)^2 * (0.5)^(3-2)
Using the binomial coefficient formula, we get:
3C2 = 3! / (2! * (3-2)!) = 3
Substituting all the values, we have:
P(2) = 3 * (0.5)^2 * (0.5)^(3-2)
= 3 * 0.25 * 0.5
= 0.375
Therefore, the probability of tossing exactly 2 heads on 3 tosses of a fair coin is 0.375 or 37.5%.