Respuesta :
Answer:
a.) 0.7063
b.) 23
Step-by-step explanation:
a.)
Let X be an event in which at least 2 students have same birthday
Y be an event in which no student have same birthday.
Now,
P(X) + P(Y) = 1
⇒P(X) = 1 - P(Y)
as we know that,
Probability of no one has birthday on same day = P(Y)
⇒P(Y) = [tex]\frac{365!}{(365)^{n} (365-n)! }[/tex] where there are n people in a group
As given,
n = 30
⇒P(Y) = [tex]\frac{365!}{(365)^{30} (365-30)! } = \frac{365!}{(365)^{30} (335)! }[/tex] = 0.2937
∴ we get
P(X) = 1 - 0.2937 = 0.7063
So,
The probability that at least two of them have their birthdays on the same day = 0.7063
b.)
Given, P(X) > 0.5
As
P(X) + P(Y) = 1
⇒P(Y) ≤ 0.5
As
P(Y) = [tex]\frac{365!}{(365)^{n} (365-n)! }[/tex]
We use hit and trial method
If n = 1 , then
P(Y) = [tex]\frac{365!}{(365)^{1} (365-1)! } = \frac{365!}{(365)^{1} (364)! }[/tex] = 1 [tex]\nleq[/tex] 0.5
If n = 5 , then
P(Y) = [tex]\frac{365!}{(365)^{5} (365-5)! } = \frac{365!}{(365)^{5} (360)! }[/tex] = 0.97 [tex]\nleq[/tex] 0.5
If n = 10 , then
P(Y) = [tex]\frac{365!}{(365)^{10} (365-10)! } = \frac{365!}{(365)^{10} (354)! }[/tex] = 0.88 [tex]\nleq[/tex] 0.5
If n = 15 , then
P(Y) = [tex]\frac{365!}{(365)^{15} (365-15)! } = \frac{365!}{(365)^{15} (350)! }[/tex] = 0.75 [tex]\nleq[/tex] 0.5
If n = 20 , then
P(Y) = [tex]\frac{365!}{(365)^{20} (365-20)! } = \frac{365!}{(365)^{20} (345)! }[/tex] = 0.588 [tex]\nleq[/tex] 0.5
If n = 22 , then
P(Y) = [tex]\frac{365!}{(365)^{22} (365-22)! } = \frac{365!}{(365)^{22} (343)! }[/tex] = 0.52 [tex]\nleq[/tex] 0.5
If n = 23 , then
P(Y) = [tex]\frac{365!}{(365)^{23} (365-23)! } = \frac{365!}{(365)^{23} (342)! }[/tex] = 0.49 [tex]\nleq[/tex] 0.5
∴ we get
Number of students should be in class in order to have this probability above 0.5 = 23
