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Question: Given the probability \(p = \dfrac{{87}}{{100}}\) that an event will not happen, how do you find the...

Given the probability p=87100p = \dfrac{{87}}{{100}} that an event will not happen, how do you find the probability that the event will happen ?

Explanation

Solution

We need to find the probability that the event E will happen using some basic probability formula. The given events in this problem are mutually exclusive and they cannot happen simultaneously. Use the formula P(E)+P(Eˉ)=1P(E) + P(\bar E) = 1
Where, P(E)P(E) is the probability that event E will occur and P(Eˉ)P(\bar E) is the probability that the event E will not occur.
Then make rearrangement in the formula to get P(E)P(E) and substitute the given values and solve the problem to get the required probability.

Complete step by step solution:
Let us first denote the given terms.
Let P(E)P(E) denotes the probability that an event E will happen.
Let P(Eˉ)P(\bar E) denote the probability that an event E will not happen.
Given P(Eˉ)=87100P(\bar E)= \dfrac{{87}}{{100}}.
Now we know that the sum of all the probabilities of all possible mutually exclusive events of an experiment is 1.
Here we note that the event E either happens or not happens. Also these are two mutually exclusive events.
Since it cannot be that the event E happens and not happens at the same time.
Hence we can say that the probability of E + the probability of not E is equal to 1.
.i.e. P(E)+P(Eˉ)=1P(E) + P(\bar E) = 1 ……(1)
In the given problem we need probability if an event E will happen.
So we make rearrangement in the equation (1) to get P(E)P(E).
We take P(Eˉ)P(\bar E) to R.H.S. in the equation (1), we get,
P(E)=1−P(Eˉ)P(E) = 1 - P(\bar E)
⇒P(E)=1−87100\Rightarrow P(E) = 1 - \dfrac{{87}}{{100}}
Now taking LCM in the R.H.S. we get,
⇒P(E)=100−87100\Rightarrow P(E) = \dfrac{{100 - 87}}{{100}}
⇒P(E)=13100\Rightarrow P(E) = \dfrac{{13}}{{100}}
Hence the probability that an event E will happen is 13100\dfrac{{13}}{{100}}.

Note:
Remember that the result of the sum of mutually exclusive probabilities is one if all the cases are considered, that is all mutually exclusive possible events.
i.e. P(E)+P(Eˉ)=1P(E) + P(\bar E) = 1.
For example, if a coin is tossed then there are two mutually exclusive events that are getting a head and tail. Hence the probability of getting head + probability of getting tail is 1.
Also note that the probability of an event E will always be in between 0 and 1.