Answer:
1) The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Step-by-step explanation:
Given : Assume there are 365 days in a year.
To find : 1) What is the probability that ten students in a class have different birthdays?
2) What is the probability that among ten students in a class, at least two of them share a birthday?
Solution :
Total outcome = 365
1) Probability that ten students in a class have different birthdays is
The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...
The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday
P(2 born on same day) = 1- P( 2 not born on same day)
![\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B365%7D%7B365%7D%5Ctimes%20%5Cfrac%7B364%7D%7B365%7D%5D)
![\text{P(2 born on same day) }=1-[\frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B364%7D%7B365%7D%5D)
The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Answer:
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Explanation:
The figure attached shows the <em>Venn diagram </em>for the given sets.
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<em><u>a) What is the probability that the number chosen is a multiple of 3 given that it is a factor of 24?</u></em>
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From the whole numbers 1 to 15, the multiples of 3 that are factors of 24 are in the intersection of the two sets: 3, 6, and 12.
There are a total of 7 multiples of 24, from 1 to 15.
Then, there are 3 multiples of 3 out of 7 factors of 24, and the probability that the number chosen is a multiple of 3 given that is a factor of 24 is:
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<em><u>b) What is the probability that the number chosen is a factor of 24 given that it is a multiple of 3?</u></em>
The factors of 24 that are multiples of 3 are, again, 3, 6, and 12. Thus, 3 numbers.
The multiples of 3 are 3, 6, 9, 12, and 15: 5 numbers.
Then, the probability that the number chosen is a factor of 24 given that is a multiple of 3 is:
Hello!!
P (Passing a qualifying examination) = 3/4
P (Being appointed if he does pass) = 3/8
P (Recieve the job) = 3/4 × 3/8 = 9/24
Good luck :)
Answer:d
Step-by-step explanation:
I see the solution in three steps.
1.) RS ⊥ ST, RS ⊥ SQ, ∠STR ≅ ∠SQR | Given
2.) RS<span>≅RS | Reflexive Property
3.) </span><span>△RST ≅ △RSQ | AAS Triangle Congruence Property</span>
