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When Conor rings Ciara’s house, the probability that Ciara answers the phone is $\frac{1}{5}$ - Leaving Cert Mathematics - Question 1 - 2017
Question 1
When Conor rings Ciara’s house, the probability that Ciara answers the phone is $\frac{1}{5}$. (a) Conor rings Ciara’s house once every day for 7 consecutive days... show full transcript
Worked Solution & Example Answer:When Conor rings Ciara’s house, the probability that Ciara answers the phone is $\frac{1}{5}$ - Leaving Cert Mathematics - Question 1 - 2017
Step 1
Find the probability that she will answer the phone on the 2$^{nd}$, 4$^{th}$, and 6$^{th}$ days but not on the other days.
Answer
To find this probability, we denote the probability of answering the phone as and the probability of not answering as . We need the probability of answering on days 2, 4, and 6, and not answering on days 1, 3, 5, and 7.
The calculation is as follows:
[ p^3 \times q^4 = \left( \frac{1}{5} \right)^3 \times \left( \frac{4}{5} \right)^4 ]
[ = \frac{1}{125} \times \frac{256}{625} = \frac{256}{78125} \approx 0.0032768 ]
Step 2
Find the probability that she will answer the phone for the 4$^{th}$ time on the 7$^{th}$ day.
Answer
To find this probability, we need to know that she has answered the phone 3 times in the first 6 days, and answers on the 7 day.
The formula for this is given by the binomial distribution:
[ P(X = 3) = {6 \choose 3} p^4 q^3 ]
Where:
- (probability of answering)
- (probability of not answering)
Now substituting the values, we have:
[ {6 \choose 3} = 20 ]
[ P(X=3) = 20 \cdot \left( \frac{1}{5} \right)^4 \cdot \left( \frac{4}{5} \right)^3 = 20 \cdot \frac{1}{625} \cdot \frac{64}{125} = \frac{1280}{78125} \approx 0.016384 ]
Step 3
Write, in terms of $n$, the probability that Ciara will answer the phone at least once.
Answer
The probability of answering at least once in days can be calculated using the complement rule.
We first calculate the probability of not answering at all:
[ P(\text{no answer in } n \text{ days}) = q^n = \left( \frac{4}{5} \right)^n ]
Thus, the probability of answering at least once is:
[ P(\text{at least one answer}) = 1 - \left( \frac{4}{5} \right)^n ]
Step 4
Find the minimum value of $n$ for which the probability that Ciara will answer the phone at least once is greater than 99%.
Answer
We set up the inequality based on the previous calculation:
[ 1 - \left( \frac{4}{5} \right)^n > 0.99 ]
Simplifying, we get:
[ \left( \frac{4}{5} \right)^n < 0.01 ]
Taking the logarithm of both sides gives:
[ n \log\left( \frac{4}{5} \right) < \log(0.01) ]
Solving for then yields:
[ n > \frac{\log(0.01)}{\log\left( \frac{4}{5} \right)} \approx 20.6377 ]
Taking the ceiling value gives us .