Having difficulty with Binomial Probabilities questionI've never worked with binomial probabilities before and I found this question below a little hard. If anyone could help that would be great.Use Binomial Probabilities to find probability that for the certain number of trials N and probability of success in each trial p number of successes will be:1) exactly k 2) less than k 3) more than kN: 10 k 6 P: 0.5Use numbers assigned to N, K, P

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Having difficulty with Binomial Probabilities questionI&#039;ve never worked with binomial probabilities before and I found this question below a little hard. If anyone could help that would be great.Use Binomial Probabilities to find probability that for the certain number of trials N and probability of success in each trial p number of successes will be:1) exactly k 2) less than k 3) more than kN: 10 k 6 P: 0.5Use numbers assigned to N, K, P

vermieterbx · Accepted Answer

Step 1Hint/partial answer: for a binomial probability where X is the number of successes and p is the probability of success,   P  (  X  =  k  )  =                              (                            N        k                              )                          p    k    (  1  −  p      )          N      −      k      .Step 2To get less than k you have to add up several of these probabilities.

Beckett Henry · Accepted Answer

Step 1The probability mass function (PMF) for a binomially distributed random variable,   X  ∼      B    i    n    (  N  ,  p  )  , is:      P    (  X  =  k  )  =                              (                            N        k                              )                          p    k    (  1  −  p      )          N      −      k          [  k  ∈  {  0..  N  }  ]From that you have the cumulative distribution function (CDF):      P    (  X  ≤  k  )  =      ∑          j      =      0        k                                (                            N        j                              )                          p    j    (  1  −  p      )          N      −      j          [  k  ∈  {  0..  N  }  ]Step 2And thusly:      P    (  X  &amp;lt;  k  )  =      ∑          j      =      0              k      −      1                                    (                            N        j                              )                          p    j    (  1  −  p      )          N      −      j          [  k  ∈  {  1..  N  }        P    (  X  &amp;gt;  k  )  =      ∑          j      =      k      +      1        N                                (                            N        j                              )                          p    j    (  1  −  p      )          N      −      j          [  k  ∈  {  0..  N  −  1  }  ]Everything else is just substituting values into the appropriate formula and calculating.

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