# The probability of getting a passing grade for an exam

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
Step 1
We know that the negative binomial is a discrete distribution whose probability mass function is given by;
$P(X=x)=(\begin{array}{c}x+r-1\\ x\end{array})p^{r}(1-p)^{x}, x=0,1,2,...$
Notation: $$\displaystyle{X}{\sin{{N}}}{B}{\left({r},{p}\right)}$$
Step 2
(4) The probability of passing an exam is 0.4.
Let $$\displaystyle{X}=$$ Number of students without a passing grade before getting a third student with a passing grade.
Then, $$\displaystyle{X}{\sin{{N}}}{B}{\left({r}={3},{p}={0.4}\right)}$$
In our problem then we need to find $$\displaystyle{P}{\left({X}={7}\right)}$$.
We need to find the probability of the 10th student is the third student with a passing grade.
$P(X=7)=(\begin{array}{c}7+3-1\\ 7\end{array})0.4^{3}(1-0.4)^{7}$
$=(\begin{array}{c}9\\7\end{array})\times 0.4^{3} \times 0.6^{7}$
$$\displaystyle={0.0645}$$
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rodclassique4r

Step 1
Let X be the number of students took the exm when we get the third student who has the passing grade.
$$\displaystyle{X}\sim{N}{e}{g}{a}{t}{i}{v}{e}\ {B}in{o}{m}{i}{a}{l}{\left({r}={3},{p}={0.4}\right)}$$
Step 2
The PMF of X is,
$P(X=x)=(\begin{array}{c}x-1\\ r-1\end{array})\cdot p^{r} \cdot (1-p)^{x-r}$
$P(X=10)=(\begin{array}{c}10-1\\ 3-1\end{array})\cdot 0.4^{3} \cdot (1-0.4)^{10-3}=(\begin{array}{c}9\\ 2\end{array})\cdot 0.4^{3} \cdot 0.6^{7}$
$$\displaystyle={36}\cdot{0.4}^{{{3}}}\cdot{0.6}^{{{7}}}$$
$$\displaystyle={0.06449725}$$

nick1337

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
$$b(x;r,p)=^{x-1}C_{r-1} \times p^{r} \times Q^{x-r}$$
$$b(10;3,0.4)=^{10-1}C_{3-1} \times 0.4^{3} \times 0.6^{10-3}$$
$$=^{9}C_{2} \times 0.4^{3} \times 0.6^{7}$$
$$=\frac{9!}{2! \times 7!} \times 0.4^{3} \times 0.6^{7}$$
$$=0.0645$$