
The probability that Frances gets a line busy whenever she calls to Michael is $\dfrac{2}{3}$ . She makes one call to Michael on each of the 4 consecutive days. Probability that the call is busy on each of the four days?
Answer
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Hint: In the given question, we need to find the probability of the given situation. Basically, by concept of probability we mean to find the possibility of occurrence of the event and then plan accordingly if any kind of situation is also given.
Complete step by step answer:
According to the question, we are given that France is someone who calls Michael and whenever he calls, he finds that line is busy with the probability of $\dfrac{2}{3}$. Also, we have been given that she does not call him daily but only on the four consecutive days. That means he calls Michael every day once and does so till four consecutive days.
Now, we need to find the probability of the situation that the call will come on all four days.
Now, the probability of a call being busy is $\dfrac{2}{3}$and similarly for every day it will remain the same as no further information is given.
So, the probability for four consecutive days is $\dfrac{2}{3}\times \dfrac{2}{3}\times \dfrac{2}{3}\times \dfrac{2}{3}=\dfrac{16}{81}$ .
Therefore, the probability of the given question is $\dfrac{16}{81}$ .
Note: We need to remember there are not any kind of restrictions mentioned while calling. We are simply asked to find the probability of the situation so we need to move around the definition of probability and then answer in spite of doing more complications and hence getting stuck.
Complete step by step answer:
According to the question, we are given that France is someone who calls Michael and whenever he calls, he finds that line is busy with the probability of $\dfrac{2}{3}$. Also, we have been given that she does not call him daily but only on the four consecutive days. That means he calls Michael every day once and does so till four consecutive days.
Now, we need to find the probability of the situation that the call will come on all four days.
Now, the probability of a call being busy is $\dfrac{2}{3}$and similarly for every day it will remain the same as no further information is given.
So, the probability for four consecutive days is $\dfrac{2}{3}\times \dfrac{2}{3}\times \dfrac{2}{3}\times \dfrac{2}{3}=\dfrac{16}{81}$ .
Therefore, the probability of the given question is $\dfrac{16}{81}$ .
Note: We need to remember there are not any kind of restrictions mentioned while calling. We are simply asked to find the probability of the situation so we need to move around the definition of probability and then answer in spite of doing more complications and hence getting stuck.
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