Help with Probablity from College

[rahulranjan86]

Hi Guys, I am new to this forum and am preparing for a competitive exam. I would be needing help with some probablity questions, essentially from college text books so that I can get my concepts clear. I would be posting questions here, and would request anybody who answers the question to please provide a explanation with the same, it would really help me understand the concept.

Thanks in advance!

Regards,
Rahul

 

[rahulranjan86]

Probability

A salesman has a 60% chance of making a sale to each customer. The behaviour of successive customers is independent. If two customers A or B enteres the shop, what is the probablity that the salesman will make a sale to A or B?

Below is my attempt:
P(Sale) = 0.6
P(A) = 0.5
P(B) = 0.5

P(S|A) = P(S and A) / P(A)
P(S|A) = 0.3/0.5 = 0.6

P(S|B) = P(S and B) / P(B)
P(S|B) = 0.3/0.5 = 0.6

after this I am not sure how to proceed rather not sure if I have approached the problem in a right manner.

 

[rahulranjan86]

Probability

If the probability that an individual suffers a bad reaction from an injection of a given serum is 0.001, determine the probability that out of 2000 individuals:
(i) exactly 3
(ii) more than 2
individuals will suffer a bad reaction.

I am really not sure how to proceed with this problem Please help!

 

[soroban]

Hello, rahulranjan86!

You are complicating the problem unnecessarily.

A salesman has a 60% chance of making a sale to each customer.
The behaviour of successive customers is independent.
Two customers, A and B, enter the shop.
What is the probablity that the salesman will make a sale to A or B?

\(\displaystyle P(\text{sells to A}) = 0.6\)
\(\displaystyle P(\text{sells to B}) = 0.6\)
\(\displaystyle P(\text{sells to A and B}) = (0.6)(0.6) = 0.36\)

\(\displaystyle \text{Therefore: }\(\text{sells to A or B}) \:=\:0.6 + 0.6 - 0.36 \:=\:0.84\)

 

[Deleted member 4993]

A salesman has a 60% chance of making a sale to each customer. The behaviour of successive customers is independent. If two customers A or B enteres the shop, what is the probablity that the salesman will make a sale to A or B?

Below is my attempt:
P(Sale) = 0.6
P(A) = 0.5
P(B) = 0.5

P(S|A) = P(S and A) / P(A)
P(S|A) = 0.3/0.5 = 0.6

P(S|B) = P(S and B) / P(B)
P(S|B) = 0.3/0.5 = 0.6

after this I am not sure how to proceed rather not sure if I have approached the problem in a right manner.

Think of it another way:

Probability of no sale to each customer is = 0.4 (independent)

Then

Probability of no sale to both customers is = 0.4² = 0.16

Then

Probability of sale to one customers is = 1 - 0.4² = 1 - 0.16 = 0.84

 

[soroban]

Hello again, rahulranjan86!

You should start another thread for a new problem.

If the probability that a person suffers a bad reaction from an injection of a given serum is 0.001,
determine the probability that out of 2000 individuals:

(1) exactly 3 have a bad reaction.

You are expected to be familiar with Binomial Probability.

We have: .\(\displaystyle \begin{Bmatrix} P(B) &=& 0.001 \\ P(B') &=& 0.999 \end{Bmatrix}\)

Therefore: .\(\displaystyle \displaystyle P(\text{3B}) \;=\;{2000\choose3}(0.001)^3(0.999)^{1997} \;=\;0.180537328\)

(2) more than 2 have a bad reaction.

Consider the opposite: "2 or less".
This means: .\(\displaystyle 0B\,\text{ or }\,1B\,\text{ or }\,2B\)

\(\displaystyle P(\text{0B}) \;=\;{2000\choose0}(0.001)^0(0.999)^{2000} \;=\; 0.135199925\)

\(\displaystyle P(\text{1B}) \;=\;{2000\choose1}(0.001)^1(0.999)^{1999} \;=\; 0.270670521\)

\(\displaystyle P(\text{2B}) \;=\;{2000\choose2}(0.001)^2(0.999)^{1998} \;=\; 0.270805992\)

Hence: .\(\displaystyle P(\text{at most 2B}) \;=\;0.676676438\)


Therefore: .\(\displaystyle P(\text{at least 3B}) \;=\;1 - 0.676676438 \;=\; 0.323323562\)

 

[rahulranjan86]

Thank You!

Thanks Guys...you guys rock! And my apologies, I will start a new thread when I come up with the new problem next time.