@havfanridindis Yes sorry I forgot the n-1 choose 2. And yes I also can't see how we will add all the infinite possible terms... There is no direct way apparently. And as you are saying the f

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## Forum (529)

Probability problems *havfanridindis *

@sp1234 In the general formula P(X=n) = (1/4)^3*( 3/4)^(n-3), didn't you forget to include all the possibilities of passing 3 and failing n-3 tests? I mean, the last test has to be one of the 3 passe

Let the no of attempts again be denotes by X X is minimum 3. If P(X=3) is 1/4 whole cubed = 1/64 P(X=4) means he passed 2 out of the first 3 and then passed the 4th. So it is (1/4)^2 * (3/4)^

Probability problems *havfanridindis *

Do I need a different formula for the expected value this time? Cause the number of possible outcomes is infinite. Problem 5: A student needs to pass 3 more courses to graduate and each time he

Probability problems *havfanridindis *

Thanks @sp1234

Problem 3 is 1-( probability that all three 4,5,6 don't occur together.) 1-{ P(not 4) + P(Not 5) + P(Not 6)- P(not 4 and not 5) - P(not 4 and not 6) - P( not 5 and not 6)+ P( not 4 and not 5 and

Problem 1 A is direct formula based. P(X=5) =1/6 P(X=2)= 1/3 P(X=-3)=1/2 Now apply the formula of mean(expected value)= 5/6 +2/3-3/2 Variance = mean of squares of X - Expected Value^2

Probability problems *havfanridindis *

Problem 4: Let X be a discrete random variable where the values for X belong to the set {2 , 3 , . . .} and the probability function: fX(x) = P( X = x ) = c [(x − 1)/x!] , x = 2, 3, . . . ,

Probability problems *havfanridindis *

Problem 3: I throw a regular dice 7 consecutive times. What's the probability of getting all 3 outcomes '4' , '5' and '6' (at least one time each)?

Probability problems *havfanridindis *

Problem 2: Let X be a random continuous variable with probability density function: fX(x)= { c/x , 1<x<2 and = 0 , otherwise} where c>0 is a constant