Question

There are \[5\% \] defective items in a large bunk of items. The probability that a sample of 10 items will include not more than 1 defective item is \[\left( {\dfrac{{{{19}^9} \times 29}}{{{{20}^x}}}} \right)\]then what is the value of x?

Answer 1

Hint:Here we use the method of random variable where selecting an item from a given number of items is the random experiment. We assume a random variable for selecting a number of defective items from the total number of items and write the general formula of probability. We take the value of success as the percentage of defective items and failure as the remaining items that are not defected. We find probability of not more than 1 defective piece by calculating the sum of probabilities of zero defective pieces and 1 defective piece.

Formula used:For a random variable X, if probability of success of an even is p and probability of loss an event is q. Then probability of choosing r success out of n total items is given by \[P(X = r){ = ^n}{C_r}{(p)^r}{(q)^{n - r}}\]
* If p is the probability of success then probability of failure q is given by \[q = 1 - p\].
* Combination formula is given by \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where n is total number of items w are choosing from and r is number of items we are selecting.

Complete step-by-step answer:
Let p denote probability of defective items from bulk items.
We are given p is \[5\% \]
We convert the percentage into fraction by dividing the number by 100.
Then, \[p = \dfrac{5}{{100}}\]
Cancel the same factors from numerator and denominator.
\[p = \dfrac{1}{{20}}\]
Then the probability of selecting an item that is not defective is q. We know that \[q = 1 - p\]
\[q = 1 - \dfrac{1}{{20}}\]
Take LCM in RHS
\[q = \dfrac{{20 - 1}}{{20}}\]
\[q = \dfrac{{19}}{{20}}\]
Let us assume the random variable X for number of defective items out of n items
Probability of r defective items out of n total items is given by \[P(X = r){ = ^n}{C_r}{(p)^r}{(q)^{n - r}}\]
We have to find the probability of defective items from a sample of 10 items. So we substitute n as 10
\[ \Rightarrow P(X = r){ = ^{10}}{C_r}{(p)^r}{(q)^{10 - r}}\]
Substitute the value of \[p = \dfrac{1}{{20}},q = \dfrac{{19}}{{20}}\]
\[ \Rightarrow P(X = r){ = ^{10}}{C_r}{(\dfrac{1}{{20}})^r}{(\dfrac{{19}}{{20}})^{10 - r}}…..(1)\]
We have to find a probability of not more than 1 item is defective.
That is the same as the sum of probabilities of zero items defective and one item defective.
Therefore, we calculate \[P(X = 0) + P(X = 1)\]
Using equation (1) we can write
\[ \Rightarrow P(X = 0){ = ^{10}}{C_0}{(\dfrac{1}{{20}})^0}{(\dfrac{{19}}{{20}})^{10 - 0}}\]
Using the formula for combination we can write\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\].
\[ \Rightarrow P(X = 0) = \dfrac{{10!}}{{(10 - 0)!0!}}{(\dfrac{1}{{20}})^0}{(\dfrac{{19}}{{20}})^{10 - 0}}\]
We know any number having power zero is equal to 1. Also, \[0! = 1\]
\[ \Rightarrow P(X = 0) = \dfrac{{10!}}{{10!}} \times 1 \times {(\dfrac{{19}}{{20}})^{10}}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow P(X = 0) = {(\dfrac{{19}}{{20}})^{10}} ……...… (2) \]
Again using equation (1)
\[ \Rightarrow P(X = 1){ = ^{10}}{C_1}{(\dfrac{1}{{20}})^1}{(\dfrac{{19}}{{20}})^{10 - 1}}\]
Using the formula for combination we can write\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\].
\[ \Rightarrow P(X = 1) = \dfrac{{10!}}{{(10 - 1)!1!}}{(\dfrac{1}{{20}})^1}{(\dfrac{{19}}{{20}})^{10 - 1}}\]
\[ \Rightarrow P(X = 1) = \dfrac{{10!}}{{9!1!}} \times (\dfrac{1}{{20}}) \times {(\dfrac{{19}}{{20}})^9}\]
Using the factorial formula \[n! = n(n - 1)!\]and put \[1! = 1\]
\[ \Rightarrow P(X = 1) = \dfrac{{10 \times 9!}}{{9!}} \times (\dfrac{1}{{20}}) \times {(\dfrac{{19}}{{20}})^9}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow P(X = 1) = 10(\dfrac{1}{{20}}){(\dfrac{{19}}{{20}})^9} . ….… (3)\]
Add equation (2) and (3)
\[ \Rightarrow P(X = 0) + P(X = 1) = {(\dfrac{{19}}{{20}})^{10}} + 10(\dfrac{1}{{20}}){(\dfrac{{19}}{{20}})^9}\]
\[ \Rightarrow P(X = 0) + P(X = 1) = {(\dfrac{{19}}{{20}})^{10}} + (\dfrac{{10}}{{20}}){(\dfrac{{19}}{{20}})^9}\]
Take common the value \[{(\dfrac{{19}}{{20}})^9}\]
\[ \Rightarrow P(X = 0) + P(X = 1) = {(\dfrac{{19}}{{20}})^9}\left[ {\dfrac{{19}}{{20}} + \dfrac{{10}}{{20}}} \right]\]
Add the terms in the bracket
\[ \Rightarrow P(X = 0) + P(X = 1) = {(\dfrac{{19}}{{20}})^9}\left[ {\dfrac{{29}}{{20}}} \right]\]
We can separate the powers of numerator and denominator.
\[ \Rightarrow P(X = 0) + P(X = 1) = \dfrac{{{{19}^9}}}{{{{20}^9}}} \times \left[ {\dfrac{{29}}{{20}}} \right]\]
Multiply the terms in numerator and denominator.
\[ \Rightarrow P(X = 0) + P(X = 1) = \dfrac{{{{19}^9} \times 29}}{{{{20}^9} \times 20}}\]
Since, we know that powers can be added when the base is the same.
\[ \Rightarrow P(X = 0) + P(X = 1) = \dfrac{{{{19}^9} \times 29}}{{{{20}^{9 + 1}}}}\]
\[ \Rightarrow P(X = 0) + P(X = 1) = \dfrac{{{{19}^9} \times 29}}{{{{20}^{10}}}}\]
Comparing the value of probability to the given value in question we get
\[ \Rightarrow \left( {\dfrac{{{{19}^9} \times 29}}{{{{20}^x}}}} \right) = \dfrac{{{{19}^9} \times 29}}{{{{20}^{10}}}}\]
We know when the base has the same powers can be equated on both sides of the equation.
\[{20^x} = {20^{10}}\]
So, \[x = 10\]
Therefore, the value of x is 10.

Note:Students make the mistake of assuming the probability of success as choosing on defective items in which case we will have to flip the formula as we will be finding the probability of failure then. Students make mistakes in writing \[0! = 0\] which makes the calculation wrong, keep in mind \[0! = 1\].