A bag contains 6 red balls and 3 blue balls. 4 balls are selected at random. find the probability of selecting 4 red balls.
1 answer:
Assuming that the balls are not replaced;
= 5/42
(the numbers are decreasing in the denominator as you are presumably taking out the red balls. In the denominator, there is one less ball each time since you took out a red one).
Therefore, the probability of selected 4 red balls consecutively without replacement is 5/42.
Hope I Helped :)
(the numbers are decreasing in the denominator as you are presumably taking out the red balls. In the denominator, there is one less ball each time since you took out a red one).
Therefore, the probability of selected 4 red balls consecutively without replacement is 5/42.
Hope I Helped :)
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Answer:
The probability that a jar contains more than 466 g is 0.119.
Step-by-step explanation:
We are given that a jar of peanut butter contains a mean of 454 g with a standard deviation of 10.2 g.
Let X = <u><em>Amount of peanut butter in a jar</em></u>
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean = 454 g
= standard deviation = 10.2 g
So, X ~ Normal()
Now, the probability that a jar contains more than 466 g is given by = P(X > 466 g)
P(X > 466 g) = P( >
) = P(Z > 1.18) = 1 - P(Z
1.18)
= 1 - 0.881 = <u>0.119</u>
The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.
Answer: 0.9862
Step-by-step explanation:
Given : The probability that the chips belongs to Japan: P(J)= 0.36
The probability that the chips belongs to United States : P(U)= 1-0.36=0.64
The proportion of Japanese chips are defective : P(D|J)=0.017
The proportion of American chips are defective : P(D|U)=0.012
Using law of total probability , we have
Thus , the probability that chip is defective = 0.0138
Then , the probability that a chip is defect-free=