Brainliest pls help (dont have to show work)
1 answer:
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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) = where there are n people in a group
As given,
n = 30
⇒P(Y) = = 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) =
We use hit and trial method
If n = 1 , then
P(Y) = = 1
0.5
If n = 5 , then
P(Y) = = 0.97
0.5
If n = 10 , then
P(Y) = = 0.88
0.5
If n = 15 , then
P(Y) = = 0.75
0.5
If n = 20 , then
P(Y) = = 0.588
0.5
If n = 22 , then
P(Y) = = 0.52
0.5
If n = 23 , then
P(Y) = = 0.49
0.5
∴ we get
Number of students should be in class in order to have this probability above 0.5 = 23