Question

If a die and a coin are thrown together, what is the probability of getting 6 and heads?

Answer 1

Here, we are asked to find the probability of getting head on coin and number 6 on die.
So, firstly, we will write down all the possible outcomes in the sample space.
Thus, count the number of outcomes in the sample space.
Let X be the event getting head on coin and number 6 on die.
Then, check how much the number of outcomes is favourable to the given event X.
Hence, find the probability of the given event using the formula $P\left( X \right) = \dfrac{{Number\,of\,outcomes\,favourable\,to\,the\,given\,event}}{{Total\,number\,of\,outcomes\,in\,the\,sample\,space}}$ .

Complete step by step solution:
The given events are throwing of a dice and tossing of a coin.
Let, H be the event where the coin shows head and T is the event where the coin shows tail.
Now, the sample space for the given events can be written as S = \left\\{ {\left( {1,H} \right),\left( {2,H} \right),\left( {3,H} \right),\left( {4,H} \right),\left( {5,H} \right),\left( {6,H} \right),\left( {1,T} \right),\left( {2,T} \right),\left( {3,T} \right),\left( {4,T} \right),\left( {5,T} \right),\left( {6,T} \right)} \right\\}
Therefore, the total number of outcomes $n\left( S \right) = 12$ .
Now, we are asked to find the probability of the coin showing head and the die showing 6 simultaneously.
Let X be the event that the coin shows head and the die shows 6.
Now, the total number of outcomes favourable to the event X is 1 which is 6 appears on die and head appears on coin i.e. $\left( {6,H} \right)$ .
$\therefore n\left( X \right) = 1$
Thus, the probability of getting head and 6 is $P\left( X \right) = \dfrac{{Number\,of\,outcomes\,favourable\,to\,the\,given\,event}}{{Total\,number\,of\,outcomes\,in\,the\,sample\,space}}$ .
$\therefore P\left( X \right) = \dfrac{{n\left( X \right)}}{{n\left( S \right)}}$
$\therefore P\left( X \right) = \dfrac{1}{{12}}$

Hence, the probability of getting number 6 on die and head on coin is $\dfrac{1}{{12}}$.

Note:
Alternate method:
Let X be the probability of getting number 6 on the die.
We know that the probability of getting any one of the numbers of die on throwing is $\dfrac{1}{6}$.
So, here $P\left( X \right) = \dfrac{1}{6}$.
Let H be the event that the coin shows heads.
Also, we know that the probability of getting a head on the coin when it is tossed is $\dfrac{1}{2}$.
So, we get $P\left( H \right) = \dfrac{1}{2}$.
Now, let A be the probability that the above events occur simultaneously.
So, to get the probability of the event A occurring can be given by the product of the occurrence of the individual events X and H.
$\therefore P\left( A \right) = P\left( X \right) \cdot P\left( H \right)$
$= \dfrac{1}{6} \times \dfrac{1}{2} \\\ = \dfrac{1}{{12}}$
Thus, the probability of getting number 6 on die and head on coin is $\dfrac{1}{{12}}$.