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.
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?
5 answers
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)?
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%.
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).
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%.