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A letter is taken out at random from the word ASSISTANT and another from STATISTICS. The probability that they are the same letters is :
A. $\dfrac{13}{90}$
B. $\dfrac{17}{90}$
C. $\dfrac{19}{90}$
D. $\dfrac{15}{90}$

Answer Verified Verified
Hint: For finding the probability, we will first the letters that represent in both of the words, ASSISTANT and STATISTICS. Then we will find out the probability that a letter is taken out from the first word and will find out the same letter from the other word. As we have to find a letter that is present in both the words, we will use the formula, $P\left( {{A}_{1}}.{{B}_{1}} \right)=P\left( {{A}_{1}} \right).P\left( {{B}_{1}} \right)$, where P (A) and P (B) are the probabilities of any two events A and B. But we have to find the probability for all the words that are common in both the words, so we have to add the probability of each word that are taken out from both the words. The formula for the probability of any event A is given as, $P\left( A \right)=\dfrac{\text{ number of favourable outcomes of A}}{\text{total number of outcomes}}$.

Complete step by step solution:
We have been given two words ASSISTANT and STATISTICS and have been asked to find the probability of taking out the same letters. We can see that the letters that are common in both the words, ASSISTANT and STATISTICS are A, S, I and T. So, we will find the probability of taking out each of these common letters in both the words one by one, using the formula for the probability of any event A as, $P\left( A \right)=\dfrac{\text{ number of favourable outcomes of A}}{\text{total number of outcomes}}$.
We will find the probability of taking letter A from the first word, ASSISTANT. There are 2 As in this word and the total number of letters is 9. So, the probability of taking letter A from the first word is $\dfrac{2}{9}$.
Now, for finding the probability of taking letter A from the second word, STATISTICS, we know that is 1 A and the total number of letters here are 10. So, the probability of taking letter A from the second word is $\dfrac{1}{10}$.
So, now, we will find the probability that letter A is taken out from both the words, using the formula, $P\left( {{A}_{1}}.{{B}_{1}} \right)=P\left( {{A}_{1}} \right).P\left( {{B}_{1}} \right)$. So, we get,
$P\left( A \right)=\dfrac{2}{9}\times \dfrac{1}{10}=\dfrac{2}{90}\ldots \ldots \ldots \left( i \right)$
Now, we will find the probability of taking letter S from the first word, ASSISTANT. There are 3 S in this word and the total number of letters is 9. So, the probability of taking letter S from the first word is $\dfrac{3}{9}$.
And, for finding the probability of taking letter S from the second word, STATISTICS, we have 3 S here and the total number of letters as 10. So, the probability of taking letter S from the second word is $\dfrac{3}{10}$.
So, we will get the probability that letter S is taken out from both the words, using the formula as,
$P\left( S \right)=\dfrac{3}{9}\times \dfrac{3}{10}=\dfrac{9}{90}\ldots \ldots \ldots \left( ii \right)$
Now we have to find the probability of taking a letter I from the first word, ASSISTANT. There is 1 I in this word and the total number of letters is 9. So, the probability of taking a letter I from the first word is $\dfrac{1}{9}$.
Now, for the probability of taking letter I from the second word, STATISTICS, we know that is 2 I and the total number of letters here are 10. So, the probability of taking a letter I from the second word is $\dfrac{2}{10}$.
So, we can find the probability that letter I am taken out from both the words, using the formula as,
$P\left( I \right)=\dfrac{1}{9}\times \dfrac{2}{10}=\dfrac{2}{90}\ldots \ldots \ldots \left( iii \right)$
And finally, we will find the probability of taking letter T from the first word, ASSISTANT. There are 2 T in this word and the total number of letters is 9. So, the probability of taking a letter T from the first word is $\dfrac{2}{9}$.
Now, for finding the probability of taking letter T from the second word, STATISTICS, we know that there are 3 T and the total number of letters are 10. So, the probability of taking a letter T from the second word is $\dfrac{3}{10}$.
So, we will find the probability that letter T is taken out from both the words, using the formula as,
$P\left( T \right)=\dfrac{2}{9}\times \dfrac{3}{10}=\dfrac{6}{90}\ldots \ldots \ldots \left( iv \right)$
Now, that we have found the probabilities of each of the common letters being taken from both the words, we can find the probability of the same letters being taken as,
$P\left( A \right)+P\left( S \right)+P\left( I \right)+P\left( T \right)$
We have these values from equations (i), (ii), (iii) and (iv). So, we will get the probability as,
$\dfrac{2}{90}+\dfrac{9}{90}+\dfrac{2}{90}+\dfrac{6}{90}=\dfrac{19}{90}$
So, the probability of the same letters being taken is $\dfrac{19}{90}$.
Therefore, the correct answer is option C.

Note: There are chances that the students might get confused with the formula and use $P\left( A.B \right)=P\left( A \right)+P\left( B \right)$ for finding the probability of any letter being taken out from both the words instead of $P\left( A.B \right)=P\left( A \right).P\left( B \right)$. But students should know when two events are independent, the probability of both occurring is the product of the probabilities of the individual events, that is, $P\left( A\text{ }and\text{ }B \right)P\left( A.B \right)=P\left( A \right).P\left( B \right)$.

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