Question

If the probability of winning a game is $\dfrac{5}{{11}}$, what is the probability of losing?

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 multiply it by a 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 the given that the probability of winning a game is $\dfrac{5}{{11}}$, which means the total event is given as $11$ and then the favorable winning game is given as $5$
Hence in the games, there are a total of two outcomes only either win or lose.
Thus, the losing probability can be calculating using the total count subtracts the given count which means $L = T - W$ where L is the losing and T is the total count, and W is the winning
Hence, we have $L = 11 - 5 \Rightarrow 6$ which is the favorable event of the losing game
Therefore, the probability can be obtained as $P = \dfrac{F}{T} \Rightarrow \dfrac{6}{{11}}$
Hence the probability of losing the game is $\dfrac{6}{{11}}$

Note:
We are also able to solve the given problem using the formula. First, let us assume the overall total probability value is $1$ (this is the most popular concept that used in the probability that the total fraction will not exceed $1$ and everything will be calculated under the number $0 - 1$ as zero is the least possible outcome and one is the highest outcome)
Hence the probability of losing the game is $1 - \dfrac{5}{{11}} = \dfrac{6}{{11}}$ (one subtracts the given winning probability to get the losing probability)