Asked by Anonymous
you are given the probabilities listed below P(A)= 0.25, P(B) = 0.30, P(C)=0.55, P(A&C) = 0.05, P(B&C)=0, P(B/A)=0.48.
A) ARE A AND B INDEPENDENT?
B) FIND P(A&B)
C) FIND P(A OR B)
D) FIND P(A/B)
E) ARE B AND C MUTUALLY EXCLUSIVE EVENTS?
F) ARE B AND C INDEPENDENT EVENTS?
A) ARE A AND B INDEPENDENT?
B) FIND P(A&B)
C) FIND P(A OR B)
D) FIND P(A/B)
E) ARE B AND C MUTUALLY EXCLUSIVE EVENTS?
F) ARE B AND C INDEPENDENT EVENTS?
Answers
Answered by
Sally
A) A and B are not independent: P(A)*P(B)= 0.25*0.30 = 0.075. This is not equal to P(A and B) = P(B|A)*P(A) = 0.48*0.25 = 0.12.
B) P(A and B) = P(B|A)*P(A) = 0.48*0.25 = 0.12.
C) P(A or B) = P(A) + P(B) - P(A and B) = 0.25 + 0.30 - 0.12 = 0.43
D) P(A|B) = P(A and B)/P(B) = 0.12/0.30 = 0.40
E) B and C are mutually exclusive, because it is given that P(B and C)=0 (Definition of mutual exclusivity.)
F) P(B)*P(C) = 0.30*0.55 = 0.165
P(B and C) = 0
==> The two quantities are not equal, hence B and C are not independent.
B) P(A and B) = P(B|A)*P(A) = 0.48*0.25 = 0.12.
C) P(A or B) = P(A) + P(B) - P(A and B) = 0.25 + 0.30 - 0.12 = 0.43
D) P(A|B) = P(A and B)/P(B) = 0.12/0.30 = 0.40
E) B and C are mutually exclusive, because it is given that P(B and C)=0 (Definition of mutual exclusivity.)
F) P(B)*P(C) = 0.30*0.55 = 0.165
P(B and C) = 0
==> The two quantities are not equal, hence B and C are not independent.
Answered by
ilyas on connexus
a bag contains 7 red marbles, 6 green marbles, 5 yellow marbles, and 2 orange marbles. marbles are drawn twice with replacement. what is p(yellow, then green)?
Answered by
Bot
The probability of drawing yellow then green is the product of the probability of drawing a yellow marble on the first draw (with replacement), and the probability of drawing a green marble on the second draw (with replacement).
The probability of drawing a yellow marble on the first draw is 5/20 (or 1/4), since there are 5 yellow marbles out of a total of 20 marbles in the bag.
The probability of drawing a green marble on the second draw is also 6/20, since there are still 6 green marbles left out of the 20 marbles (assuming the first marble was replaced).
Therefore, the probability of drawing yellow then green is (5/20) * (6/20) = 0.075 or 7.5%.
The probability of drawing a yellow marble on the first draw is 5/20 (or 1/4), since there are 5 yellow marbles out of a total of 20 marbles in the bag.
The probability of drawing a green marble on the second draw is also 6/20, since there are still 6 green marbles left out of the 20 marbles (assuming the first marble was replaced).
Therefore, the probability of drawing yellow then green is (5/20) * (6/20) = 0.075 or 7.5%.
Answered by
ilyas on connexus
letter tiles that spell the word happiness are placed in a bag. A tile is drawn and replaced and then a second tile is drawn. find p(vowel, then p).
Answered by
Bot
There are 3 vowels (a, e, i) and 7 consonants (h, n, p, s) in the word happiness, for a total of 10 tiles.
The probability of drawing a vowel on the first draw (with replacement) is 3/10, since there are 3 vowels out of 10 tiles in the bag.
The probability of drawing a consonant on the second draw (with replacement) is 7/10, since there are 7 consonants left out of the 10 tiles (assuming the first tile was replaced).
Therefore, the probability of drawing a vowel then a consonant is (3/10) * (7/10) = 0.21 or 21%.
The probability of drawing a vowel on the first draw (with replacement) is 3/10, since there are 3 vowels out of 10 tiles in the bag.
The probability of drawing a consonant on the second draw (with replacement) is 7/10, since there are 7 consonants left out of the 10 tiles (assuming the first tile was replaced).
Therefore, the probability of drawing a vowel then a consonant is (3/10) * (7/10) = 0.21 or 21%.