The probability of getting a passing grade for an exam is 0.4. What is the probability of getting the third student who has a passing grade when 10th student have took the exam

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raefx88y · Accepted Answer

Step 1  We know that the negative binomial is a discrete distribution whose probability mass function is given by;  P(X=x)=(x+r−1x)pr(1−p)x,x=0,1,2,...  Notation: Xsin⁡NB(r,p)  Step 2  (4) The probability of passing an exam is 0.4.  Let X= Number of students without a passing grade before getting a third student with a passing grade.  Then, Xsin⁡NB(r=3,p=0.4)  In our problem then we need to find P(X=7).  We need to find the probability of the 10th student is the third student with a passing grade.  P(X=7)=(7+3−17)0.43(1−0.4)7  =(97)×0.43×0.67  =0.0645

rodclassique4r · Accepted Answer

Step 1 Let X be the number of students took the exm when we get the third student who has the passing grade. X∼Negative Binomial(r=3,p=0.4) Step 2 The PMF of X is, P(X=x)=(x−1r−1)⋅pr⋅(1−p)x−r P(X=10)=(10−13−1)⋅0.43⋅(1−0.4)10−3=(92)⋅0.43⋅0.67 =36⋅0.43⋅0.67 =0.06449725

nick1337 · Accepted Answer

Step 1 Given, The probability of getting passing grade in an exam is p=0.4 Therefore the probability of not getting passing grade in an exam is obtained as Q=1-0.4=0.6 The number of success r=3 The number of trials required to produce r success is x=10 Step 2 Therefore, the required probability of getting the third student who has a passing grade when 10th student have took  the exam is obtained as b(x;r,p)=x−1Cr−1×pr×Qx−r b(10;3,0.4)=10−1C3−1×0.43×0.610−3 =9C2×0.43×0.67 =9!2!×7!×0.43×0.67 =0.0645

The probability of getting a passing grade for an exam

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