If we randomly pick two persons from the population, each one will have a probability of 1-1/28=27/28 of being a non-carrier.
Thus for neither to be non-carrier, the probability is the product of the individual probabilities since the picks are assumed independent (with a large enough population).
Probability = (27/28)*(27/28) = 729/784
or approximately 0.930
The probability of winning exactly 21 times is 0.14 when the probability of winning the arcade game is 0.659.
We know that binomial probability is given by:
Probability (P) = ⁿCₓ (probability of 1st)ˣ x (1 - probability of 1st)ⁿ⁻ˣ
We are given that:
Probability of winning on an arcade game = P(A) = 0.659
So, the Probability of loosing on an arcade game will be = P'(A) = 1 - 0.659 = 0.341
Number of times the game is being played = 30
We have to find the Probability of winning exactly 21 times.
Here,
n = 30
x = 21
P(A) = 0.659
P'(A) = 0.341
Using the binomial probability formula, we get that:
Probability of winning exactly 21 times :
P(21 times) = ³⁰C₂₁ (0.659)²¹ x (0.341)⁷
P( 21 times ) = 0.14
Therefore, the probability of winning exactly 21 times is 0.14
Learn more about " Binomial Probability " here: brainly.com/question/12474772
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Answer:
first and third one
Step-by-step explanation:
if x=0 then it is a y intercept. if y=0 then it is an x intercept. if both then it is both
Answer:
C it wasnt hard
Step-by-step explanation:
Meredith wrote an algebraic expression because it has two unknown variables.
