1. 1/4
2. 2/4
3. 1/6
4. 3/4
Probability of Getting At Least One Head in Two Coin Flips
This problem asks for the probability of a specific outcome when flipping a balanced coin two times. A balanced coin means that the chance of landing on heads (H) is equal to the chance of landing on tails (T), which is $1/2$ for each flip.
Understanding the Sample Space
First, let's list all the possible outcomes when flipping a coin twice. The set of all possible outcomes is called the sample space. Each flip is independent.
| Flip 1 | Flip 2 | Outcome |
| H | H | HH |
| H | T | HT |
| T | H | TH |
| T | T | TT |
There are 4 possible outcomes in total. Since the coin is balanced and the flips are independent, each of these outcomes is equally likely. The probability of each specific outcome (like HH, HT, TH, TT) is:
$P(\text{any specific outcome}) = P(\text{outcome of flip 1}) \times P(\text{outcome of flip 2}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Calculating Probability of At Least One Head
The question asks for the probability of getting "at least one head". This means we are interested in outcomes where there is one head or two heads. Looking at our sample space, the outcomes that satisfy this condition are:
- HH (Two heads)
- HT (One head)
- TH (One head)
There are 3 outcomes that have at least one head.
To find the probability of getting at least one head, we sum the probabilities of these favorable outcomes:
$P(\text{at least one head}) = P(\text{HH}) + P(\text{HT}) + P(\text{TH})$
$P(\text{at least one head}) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$
Alternative Method: Using Complementary Probability
Another way to solve this is to consider the complementary event: the event of *not* getting at least one head. This means getting *no* heads at all, which is the outcome TT (two tails).
The probability of getting two tails is:
$P(\text{TT}) = \frac{1}{4}$
The probability of an event happening is 1 minus the probability of the event not happening.
$P(\text{at least one head}) = 1 - P(\text{no heads})$
$P(\text{at least one head}) = 1 - P(\text{TT})$
$P(\text{at least one head}) = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$
Both methods confirm that the probability of getting at least one head when flipping a balanced coin twice is $3/4$. This corresponds to the fourth option.
Important Questions from Probability (Notes)
- In a box there are 4 white balls and 6 black balls. A ball is drawn at random. If it is white, it is put back along with two more white balls in the box. If it is black, it is put back in the box and then two black balls are thrown out of the box. Now a ball is drawn again at random from the box. Then, what is the probability that it is black?
- A fair coin is tossed three times. Let A be the event of getting exactly two heads and B be the event of getting at most
two tails, then P(A$\cup$B) is: Bag A contains 3 Red and 4 Black balls while Bag B contains 5 Red and 6 Black balls. One ball is drawn at random from one of the bags and is found to be red. Then, the probability that it was drawn from Bag B is
Suppose that the random variable X takes on the values: -1, 0, and 2 with probability $\frac{1}{8}$, $\frac{1}{2}$ and $\frac{3}{8}$. Find the expected value of X.
If the probability function for a random variable x is given as f(x) = (x+3)/15 when x = 1, 2 and 3. Find the sum of the values of the probability distribution for x.
