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
\[0.25\].
Events and
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events. More formally, if events and are independent, then the probability of both and occurring is:
where is the probability of events and both occurring, P(A) is the probability of event occurring, and is the probability of event occurring.
If you flip a coin twice, what is the probability that it will come up heads both times? Event is that the coin comes up heads on the first flip and Event is that the coin comes up heads on the second flip. Since both and equal 1/2, the probability that both events occur is
.
Let's take another example. If you flip a coin and roll a six-sided die, what is the probability that the coin comes up heads and the die comes up 1? Since the two events are independent, the probability is simply the probability of a head (which is 1/2) times the probability of the die coming up 1 (which is 1/6). Therefore, the probability of both events occurring is 1/2 x 1/6 = 1/12.
One final example: You draw a card from a deck of cards, put it back, and then draw another card. What is the probability that the first card is a heart and the second card is black? Since there are 52 cards in a deck and 13 of them are hearts, the probability that the first card is a heart is 13/52 = 1/4. Since there are 26 black cards in the deck, the probability that the second card is black is 26/52 = 1/2. The probability of both events occurring is therefore 1/4 x 1/2 = 1/8.
See the section on conditional probabilities on this page to see how to compute when A and B are not independent.
If Events A and B are independent, the probability that either Event A or Event B occurs is:
In this discussion, when we say " or occurs" we include three possibilities:
- occurs and does not occur
- occurs and does not occur
- Both and occur
This use of the word "or" is technically called inclusive or because it includes the case in which both and occur. If we included only the first two cases, then we would be using an exclusive or.
The simple event can happen if either -and- happens or -and-not- happens. Similarly, the simple event happens if either -and- happens or not--and- happens. is therefore -and- -and-not- -and- not--and- , whereas -or- is -and- -and-not) not--and- . We can make these two sums equal by subtracting one occurrence of -and- from the first. Hence, or-and-).
Now for some examples. If you flip a coin two times, what is the probability that you will get a head on the first flip or a head on the second flip (or both)? Letting Event A be a head on the first flip and Event B be a head on the second flip, then , , and . Therefore,.
If you throw a six-sided die and then flip a coin, what is the probability that you will get either a 6 on the die or a head on the coin flip (or both)? Using the formula,
An alternate approach to computing this value is to start by computing the probability of not getting either a 6 or a head. Then subtract this value from 1 to compute the probability of getting a 6 or a head. Although this is a complicated method, it has the advantage of being applicable to problems with more than two events. Here is the calculation in the present case. The probability of not getting either a 6 or a head can be recast as the probability of
(not getting a 6) AND (not getting a head).
This follows because if you did not get a 6 and you did not get a head, then you did not get a 6 or a head. The probability of not getting a six is 1 - 1/6 = 5/6. The probability of not getting a head is 1 - 1/2 = 1/2. The probability of not getting a six and not getting a head is 5/6 x 1/2 = 5/12. This is therefore the probability of not getting a 6 or a head. The probability of getting a six or a head is therefore (once again) 1 - 5/12 = 7/12.
If you throw a die three times, what is the probability that one or more of your throws will come up with a 1? That is, what is the probability of getting a 1 on the first throw OR a 1 on the second throw OR a 1 on the third throw? The easiest way to approach this problem is to compute the probability of
NOT getting a 1 on the first throw
AND not getting a 1 on the second throw
AND not getting a 1 on the third throw.
The answer will be 1 minus this probability. The probability of not getting a 1 on any of the three throws is 5/6 x 5/6 x 5/6 = 125/216. Therefore, the probability of getting a 1 on at least one of the throws is 1 - 125/216 = 91/216.