Question

A fair coin is tossed four times. What is the probability that at most three tails occur?
$A)\dfrac{7}{8}$
$B)\dfrac{{15}}{{16}}$
$C)\dfrac{{13}}{{16}}$
$D)\dfrac{3}{4}$

Answer 1

Hint: Probability is the term mathematically with events that occur, which is the number of favorable events that divides the total number of the outcomes.
If we divide the probability and then multiplied with the hundred then we will determine its percentage value.
 $\dfrac{1}{6}$ which means the favorable event is $1$ and the total outcome is $6$
 Formula used:
$P = \dfrac{F}{T}$where P is the overall probability, F is the possible favorable events and T is the total outcomes from the given.

Complete step by step answer:
Since from given that a coin is tossed four times and then the outcome is ${4^2}$ . in general, if the coin is tossed n-times and hence the total outcome is ${n^2}$
Thus, the total outcome of the probability event is $16$.
Now to find the favorable event, the probability that at most three tails occur. At most means it should not exceed the value, that is it should not exceed the three tails in the four-times tossed the coin.
Hence the possible ways are $HHHH$ (zero tails), $THHH,HTHH,HHTH,HHHT$ (one tail), $HHTT,HTTH,TTHH,HTHT,THTH,THHT$ (two tails), $TTTH,TTHT,THTT,HTTT$ (three tails)
Thus, it will not exceed three tails so $TTTT$ is not possible.
Thus, the favorable event count is $15$
Hence, we get $P = \dfrac{F}{T} \Rightarrow \dfrac{{15}}{{16}}$ is the probability that at most three tails occur.

So, the correct answer is “Option B”.

Note:
In a coin, there are only two possibilities, either head or tail. Hence in general we used the coin tossed formula as to ${n^2}$ where n is the number of times the coin is tossed.
If the coin is tossed three times then we get the total outcome as ${n^2} = {3^2} = 9$ ways.
Thus, we used the concept and said that the total outcome of the probability event is $16$ for a four-time coin toss.