Answer
Hint:
Here, we need to find the probability of getting two heads when a fair coin is tossed twice. First, we will write the sample space with all possible outcomes. Then, find the number of outcomes in the sample space where two heads are obtained. Finally, we will use the formula of probability to get the required probability.
Formula Used: The probability of an event is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Number of total
outcomes}}}}\].
Complete step by step solution:
The probability of an event is the chance that the event occurs.
We will start solving the question by writing the sample space when a fair coin is tossed twice.
We know that when a coin is tossed once, there are only two outcomes, heads or tails.
Therefore, when a coin is tossed twice, the possible outcomes are \[{\text{HH, HT, TH, TT}}\], where \[{\text{H}}\] represents heads and \[{\text{T}}\] represents tails. Here, the
first letter denotes the outcome on the first toss, and the second letter denotes the outcome on the second toss.
Thus, the sample space is \[S = \left\{ {{\text{HH, HT, TH, TT}}} \right\}\].
As there are four possible outcomes in the sample space, hence, the number of total outcomes is 4.
Next, we will find the number of times 2 heads are obtained when a fair coin is tossed twice.
We can observe that two heads appear only in the outcome \[{\text{HH}}\] in the sample space \[S =
\left\{ {{\text{HH, HT, TH, TT}}} \right\}\].
Thus, the number of favourable outcomes is 1.
Now, we need to find the probability that two heads are obtained when a fair coin is tossed twice.
Next, we know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Number of total outcomes}}}}\].
Let \[E\] be the event of getting two heads when a fair coin is tossed twice.
Substituting 1 for the number of
favourable outcomes and 4 for the number of total outcomes, we get
\[P\left( E \right) = \dfrac{1}{4}\]
\[\therefore\] The probability of getting two heads when a fair coin is tossed twice is
\[\dfrac{1}{4}\] or \[0.25\].
Note:
We can also solve this problem using the fact that the outcome in the first toss of the coin does not affect the outcome in the second toss of the coin. This means that the first toss and second toss are independent events.
We know
that when a fair coin is tossed, there are only two outcomes, heads or tails.
Let \[{\text{H}}\] represent heads and \[{\text{T}}\] represents tails.
The probability of getting a heads on the first toss of the coin is
\[P\left( H \right) = \dfrac{1}{2}\]
Similarly, the probability of getting a heads on the second toss of the coin is
\[P\left( H \right) = \dfrac{1}{2}\]
Since the first toss and second toss are independent events, we can calculate the probability of getting two
heads on two tosses of a coin by multiplying the probability of getting a heads on the first toss of the coin and the probability of getting a heads on the second toss of the coin.
Let \[E\] be the event of getting two heads when a fair coin is tossed twice.
Thus, we get
\[\begin{array}{l}P\left( E \right) = \dfrac{1}{2} \times \dfrac{1}{2}\\ = \dfrac{1}{4}\end{array}\]
\[\therefore\] The probability of getting two heads when a fair coin is tossed twice is \[\dfrac{1}{4}\] or
\[0.25\].
You are confusing the terms "independent" and "mutually exclusive". These are not the same thing. In fact events cannot be both "independent" and "mutually exclusive". It's either one, the other, or neither.
"Mutually exclusive" simply means that the two events cannot happen together. If A happens then B does not and if B happens A does not.
"Independent" simply means that the occurrence of one event is not conditional on the occurrence of the other. The probability of A happening does not depend on whether B happens or not, and vice versa.
Let $H_n$ be the indexed event of getting a head on the $n^{th}$ flip.
Given an unbiased coin, $P(H_1)=P(H_2)=\frac 1 2$
These events are independent so $P(H_1 \cap H_2) = P(H_1)\times P(H_2)$. The outcome of one coin toss does not influence the outcome of the other.
However they are not mutually exclusive, so $P(H_1 \cup H_2) = P(H_1)+P(H_2) - P(H_1 \cap H_2)$. Both coins can turn up heads.
Putting it together: $$\therefore P(H_1 \cup H_2) = \frac 12 + \frac 1 2 - \frac 12 \times \frac 12 = \frac 3 4$$
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