Introduction to Probability with replacement :
To calculate the probability of ball which is strained from a population by using explanation of probability and the calculations can be established by using the relative frequency definitions of probability. In a examination can have n equally likely conclusion, and that a 'success' can take place in s ways (from the n).
Then the probability of a 'success' = s / n
Probability with replacement Problem 1
A pitcher contains 7 orange and 4 green balls. Find the matching probabilities if the balls are replaced after every draw.
(a) Mutually orange
(b) a orange and a green
(c) both the identical color.
Solution:-
a.) Since the balls are restored after each draw,
P(s)is 7 in both cases since we replace it with in each case.
This is simply 7/11 * 7/11 = 49/121
b.) This should be 7/11 * 4/11 = 28/121
c.) As for P(c).
P(c) = (P(two orange and two green)) .,
so
P(two orange or two green) = P(two orange) + P(two green) = P(c)
P(two orange) is given by 49 / 121
P(two green) is given by 16 / 121
now by just plugging in it in the given function we get it as follows
= 49/121 + 16/121 = 65/121.
65 / 121 is the finalised answer. Please express your views of this topic Example of Theoretical Probability by commenting on blog.
Probability with replacement Problem 2
A population of 50 red mice, 200 green mice, selections with replacement:
a) Probability of 3 red mice in 3 selections
b) Probability of selecting, in order, red, red and then Green.
c) If, however, we are not interested in the order (i.e. red, red, green) but just the overall outcome (i.e. 2 red, 1 green), the probability is different:0
Solution:-
a) Probability of 3 red mice in 3 selections = (50/250) * (50/250) * (50/250)
= (1/5) * (1/5) * (1/5) = 0.008
b) Probability of selecting, in order, red, red and then green = (50/250) * (50/250) * (200/250)
= (1/5) * (1/5) * (4/5) = 0.032
c) If, however, we are not interested in the order (i.e. red, red, green) but just the overall outcome (i.e. 2 red, 1 green), the probability is different:0
Possible outcome of 3 selections with replacement
To calculate the probability of ball which is strained from a population by using explanation of probability and the calculations can be established by using the relative frequency definitions of probability. In a examination can have n equally likely conclusion, and that a 'success' can take place in s ways (from the n).
Then the probability of a 'success' = s / n
Probability with replacement Problem 1
A pitcher contains 7 orange and 4 green balls. Find the matching probabilities if the balls are replaced after every draw.
(a) Mutually orange
(b) a orange and a green
(c) both the identical color.
Solution:-
a.) Since the balls are restored after each draw,
P(s)is 7 in both cases since we replace it with in each case.
This is simply 7/11 * 7/11 = 49/121
b.) This should be 7/11 * 4/11 = 28/121
c.) As for P(c).
P(c) = (P(two orange and two green)) .,
so
P(two orange or two green) = P(two orange) + P(two green) = P(c)
P(two orange) is given by 49 / 121
P(two green) is given by 16 / 121
now by just plugging in it in the given function we get it as follows
= 49/121 + 16/121 = 65/121.
65 / 121 is the finalised answer. Please express your views of this topic Example of Theoretical Probability by commenting on blog.
Probability with replacement Problem 2
A population of 50 red mice, 200 green mice, selections with replacement:
a) Probability of 3 red mice in 3 selections
b) Probability of selecting, in order, red, red and then Green.
c) If, however, we are not interested in the order (i.e. red, red, green) but just the overall outcome (i.e. 2 red, 1 green), the probability is different:0
Solution:-
a) Probability of 3 red mice in 3 selections = (50/250) * (50/250) * (50/250)
= (1/5) * (1/5) * (1/5) = 0.008
b) Probability of selecting, in order, red, red and then green = (50/250) * (50/250) * (200/250)
= (1/5) * (1/5) * (4/5) = 0.032
c) If, however, we are not interested in the order (i.e. red, red, green) but just the overall outcome (i.e. 2 red, 1 green), the probability is different:0
Possible outcome of 3 selections with replacement
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