To determine the probability of the spinner landing on B and then C, we need to know the individual probabilities of landing on B and C. Assuming the spinner is fair and has an equal number of sections for A, B, and C, the probability of landing on B is 1/3, and the probability of landing on C is also 1/3. Thus, the combined probability of landing on B first and then C is (1/3) * (1/3) = 1/9.
The Probability of NOT reading newspaper a is .8 The Probability of NOT reading Newspaper b is .84 The probability of NOT reading Newspaper c is .86 Therefore, .8*.84*.86=0.57792=57.792%
The probability of landing on A in one spin is ( \frac{1}{4} ). To find the probability of landing on A twice in a row, you multiply the probabilities of each independent event: ( \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} ). Therefore, the probability of landing on A twice in a row is ( \frac{1}{16} ).
There are 34 ie 81 possibilities: B appears exactly twice in 18 of these: AABB, ABAB, ABBA, ABBC,ABCB, ACBB, BABA, BABC, BBAA, BBAC, BCAB, BCCB, CABB,CBAB, CBBA, CBBC, CCBA and CCBB.B comes up exactly twice in 18 of the 81 possibilities so probability = 18/81 = 0.22 or 22.2%
The probability density function for a beta distribution with parameters a and b (a, b > 0) is of the form:B(x; a, b) = c*x*(a-1)*(1-x)(b-1) where 0 <= x <= 1. c is a constant such that the integral is 1 and it can be shown that c = G(a+b)/[G(a)*G(b)] where G is the Gamma function.
The total number of alphabets is 26. So the probability of letter C = No of time c is present in the alphabets / Total number of alphabets So probability of letter c is 1/26
the probability would be 1/3
b is incorrect while c is virtually meaningless.
The Probability of NOT reading newspaper a is .8 The Probability of NOT reading Newspaper b is .84 The probability of NOT reading Newspaper c is .86 Therefore, .8*.84*.86=0.57792=57.792%
The probability of landing on A in one spin is ( \frac{1}{4} ). To find the probability of landing on A twice in a row, you multiply the probabilities of each independent event: ( \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} ). Therefore, the probability of landing on A twice in a row is ( \frac{1}{16} ).
There are 34 ie 81 possibilities: B appears exactly twice in 18 of these: AABB, ABAB, ABBA, ABBC,ABCB, ACBB, BABA, BABC, BBAA, BBAC, BCAB, BCCB, CABB,CBAB, CBBA, CBBC, CCBA and CCBB.B comes up exactly twice in 18 of the 81 possibilities so probability = 18/81 = 0.22 or 22.2%
60
It is 0.25
The answer requires more information about the context.
The probability density function for a beta distribution with parameters a and b (a, b > 0) is of the form:B(x; a, b) = c*x*(a-1)*(1-x)(b-1) where 0 <= x <= 1. c is a constant such that the integral is 1 and it can be shown that c = G(a+b)/[G(a)*G(b)] where G is the Gamma function.
Let X and Y be two random variables.Case (1) - Discrete CaseIf P(X = x) denotes the probability that the random variable X takes the value x, then the joint probability of X and Y is P(X = x and Y = y).Case (2) - Continuous CaseIf P(a < X < b) is the probability of the random variable X taking a value in the real interval (a, b), then the joint probability of X and Y is P(a < X< b and c < Y < d).Basically joint probability is the probability of two events happening (or not).
The total number of alphabets is 26. So the probability of letter C = No of time c is present in the alphabets / Total number of alphabets So probability of letter c is 1/26
(b b b)( b b b )(b d g a)(b....)(c c c c)(c b b b)(a a a b)(a...d)(b b b)(b b b)(b d g a)(b....)(c c c c)(c b b b)(d d c a)(g.....)