Part (a): Applying to 5 jobs, all five say yes
We have n = 5 as the sample size, p = 0.70 as the probability of succcess, and x = 5 as the number of successes we want to achieve. Here we define a "success" as "the company wants to hire you".
Each trial is independent of one another, we have a fixed number of trials, and the value of p stays the same for each trial. We can use the binomial probability formula
B(x) = (n C x)*(p)^x*(1-p)^(n-x)
B(x) = (5 C x)*(0.70)^x*(1-0.70)^(5-x)
B(x) = (5 C x)*(0.70)^x*(0.30)^(5-x)
The n C x notation refers to the nCr combination formula
Simply replace r with x and you have the same idea.
Two nice properties of this formula are
Meaning that the "n C x" coefficient won't really play a factor if x = 0 or x = n, since multiplying 1 with any number does not change the number.
Plug in x = 5
B(x) = (5 C x)*(0.70)^x*(0.30)^(5-x)
B(5) = (5 C 5)*(0.70)^5*(0.30)^(5-5)
B(5) = (0.70)^5
B(5) = 0.16807
You have a 16.807% chance of getting all five companies wanting to hire you.
Answer: 0.16807
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Part (b): Applying to 3 jobs, all three say no
We now have n = 3 and x = 0, so,
B(x) = (n C x)*(p)^x*(1-p)^(n-x)
B(0) = (3 C 0)*(0.70)^0*(1-0.70)^(3-0)
B(0) = (1-0.70)^(3-0)
B(0) = (0.30)^3
B(0) = 0.027
There's a 2.7% chance of not getting any positive replies.
Answer: 0.027
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Part (c): Applying to 4 jobs, at least one says yes.
Follow the same idea as part (b), but now n = 4 and x = 0
B(x) = (n C x)*(p)^x*(1-p)^(n-x)
B(0) = (4 C 0)*(0.70)^0*(1-0.70)^(4-0)
B(0) = 0.0081
There's an 0.81% chance of not getting any positive replies.
Subtract B(0) from 1 to find the probability B( x >= 0). The notation x >= 0 means "x is greater than or equal to 0".
This works because the events "getting no offers" and "getting at least one offer" are complementary events. One or the other must happen.
Put another way
P(getting no offers) + P(getting at least one offer) = 1
B(0) + B(x >= 0) = 1
B(x >= 0) = 1 - B(0)
B(x >= 0) = 1 - 0.0081
B(x >= 0) = 0.9919
There's a 99.19% chance of you getting at least one acceptance.
Answer: 0.9919
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Part (d): Determining how many applications to be at least 95% certain of getting at least one offer
Follow the same steps as part (c). We'll be doing multiple rounds of such.
We will make x = 0 be a fixed value while the variable n will progress through the positive integers {1, 2, 3, ...}. We stop the process when B(x >= 0) is 0.95 or larger.
For n = 1
B(x) = (n C x)*(p)^x*(1-p)^(n-x)
B(0) = (1 C 0)*(0.70)^0*(1-0.70)^(1-0)
B(0) = 0.3
B(x >= 0) = 1 - B(0)
B(x >= 0) = 1 - 0.3
B(x >= 0) = 0.7
After applying to 1 job, the chances of getting at least one offer is 70%
For n = 2
B(x) = (n C x)*(p)^x*(1-p)^(n-x)
B(0) = (2 C 0)*(0.70)^0*(1-0.70)^(2-0)
B(0) = 0.09
B(x >= 0) = 1 - B(0)
B(x >= 0) = 1 - 0.09
B(x >= 0) = 0.91
After applying to 2 jobs, the chances of getting at least one offer is 91%
For n = 3
B(x) = (n C x)*(p)^x*(1-p)^(n-x)
B(0) = (3 C 0)*(0.70)^0*(1-0.70)^(3-0)
B(0) = 0.027
B(x >= 0) = 1 - B(0)
B(x >= 0) = 1 - 0.027
B(x >= 0) = 0.973
After applying to 3 jobs, the chances of getting at least one offer is 97.3%
We stop here since 0.973 is larger than 0.95
If you want to be at least 95% certain of getting at least one positive result, then the lowest amount of effort to do is applying for n = 3 companies.
For any larger value of n, the value of B(x >= 0) will be larger than 0.973 but will not equal 1
So we can say 0.973 < r < 1 where r is the B(x >= 0) value and n > 3
Answer: 3 job applications (minimum)