Question

[6+4=10 points] Problem 2. Suppose that there are k people in a party with the following PMF: • k = 5 with probability 1 4 • k = 10 with probability 1 4 • k = 15 with probability 1 2 1) What is the probability that at least two people were born in the same month? Assume that all months are equally likely. Make sure to provide all decimal places in the answer. 2) Given that there are at least two people that celebrate their birthdays in the same month, what is the probability that k = 10?. Make sure to provide all decimal places in the answer.

Answer 1

Answer:

1). 0.903547

2). 0.275617

Step-by-step explanation:

It is given :

K people in a party with the following :

i). k = 5 with the probability of $\frac{1}{4}$

ii). k = 10 with the probability of $\frac{1}{4}$

iii). k = 10 with the probability $\frac{1}{2}$

So the probability of at least two person out of the 'n' born people in same month is  = 1 - P (none of the n born in the same month)

= 1 - P (choosing the n different months out of 365 days) = $1-\frac{_{n}^{12}\textrm{P}}{12^2}$

1). Hence P(at least 2 born in the same month)=P(k=5 and at least 2 born in the same month)+P(k=10 and at least 2 born in the same month)+P(k=15 and at least 2 born in the same month)

= $\frac{1}{4}\times (1-\frac{_{5}^{12}\textrm{P}}{12^5})+\frac{1}{4}\times (1-\frac{_{10}^{12}\textrm{P}}{12^{10}})+\frac{1}{2}\times (1-\frac{_{15}^{12}\textrm{P}}{12^{15}})$

= $0.25 \times 0.618056 + 0.25 \times 0.996132 + 0.5 \times 1$

= 0.903547

2).P( k = 10|at least 2 share their birthday in same month)

=P(k=10 and at least 2 born in the same month)/P(at least 2 share their birthday in same month)

= $0.25 \times \frac{0.996132}{0.903547}$

= 0.0.275617