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Question

Answers

$

{\text{A}}{\text{. }}\dfrac{7}{{144}} \\

{\text{B}}{\text{. }}\dfrac{5}{{744}} \\

{\text{C}}{\text{. }}\dfrac{7}{{544}} \\

{\text{D}}{\text{. }}\dfrac{5}{{372}} \\

$

Answer

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Hint- Here, we will be using the general formula for probability of occurrence of an event.

In a chess board, there are total 64 squares and out of these 64 squares we have to choose 3 squares. So, here combination will be used as we have to pick or choose some items from a bulk. Also, there are total 16 squares which comes in a diagonal line. Here, we have to choose 3 squares and find the probability that these 3 squares lie in a diagonal line.

As we know that, \[{\text{Probability of a event}} = \dfrac{{{\text{Possible number of outcomes }}}}{{{\text{Total number of outcomes}}}}\]

Here, possible number of outcomes will be equal to the number of ways of choosing 3 squares out of 16 squares which are present in a diagonal line.

i.e., Possible number of outcomes is \[{}^{16}{C_3}\].

Also, total number of outcomes will be equal to the number of ways of choosing 3 squares out of total 64 squares which are present in a chess box.

i.e., Total number of outcomes is \[{}^{64}{C_3}\] where \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Therefore, Required Probability\[ = \dfrac{{{}^{16}{C_3}}}{{{}^{64}{C_3}}}{\text{ = }}\dfrac{{\dfrac{{16!}}{{3!\left( {16 - 3} \right)!}}}}{{\dfrac{{64!}}{{3!\left( {64 - 3} \right)!}}}} = \dfrac{{16!3!\left( {64 - 3} \right)!}}{{64!3!\left( {16 - 3} \right)!}} = \dfrac{{16!61!}}{{64!13!}} = \dfrac{{16.15.14.13!61!}}{{64.63.62.61!13!}} = \dfrac{{16.15.14}}{{64.63.62}} = \dfrac{5}{{372}}\]

Therefore, the chance that the chosen three squares should be in a diagonal line is \[\dfrac{5}{{372}}\].

Hence, option D is correct.

Note- In these type of problems, the chance or the probability can be computed by applying the general formula for probability which is further solved by knowing the possible and total number of outcomes. Here, we are choosing that’s why taking combinations but if we had to arrange then we would have taken permutations.

In a chess board, there are total 64 squares and out of these 64 squares we have to choose 3 squares. So, here combination will be used as we have to pick or choose some items from a bulk. Also, there are total 16 squares which comes in a diagonal line. Here, we have to choose 3 squares and find the probability that these 3 squares lie in a diagonal line.

As we know that, \[{\text{Probability of a event}} = \dfrac{{{\text{Possible number of outcomes }}}}{{{\text{Total number of outcomes}}}}\]

Here, possible number of outcomes will be equal to the number of ways of choosing 3 squares out of 16 squares which are present in a diagonal line.

i.e., Possible number of outcomes is \[{}^{16}{C_3}\].

Also, total number of outcomes will be equal to the number of ways of choosing 3 squares out of total 64 squares which are present in a chess box.

i.e., Total number of outcomes is \[{}^{64}{C_3}\] where \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Therefore, Required Probability\[ = \dfrac{{{}^{16}{C_3}}}{{{}^{64}{C_3}}}{\text{ = }}\dfrac{{\dfrac{{16!}}{{3!\left( {16 - 3} \right)!}}}}{{\dfrac{{64!}}{{3!\left( {64 - 3} \right)!}}}} = \dfrac{{16!3!\left( {64 - 3} \right)!}}{{64!3!\left( {16 - 3} \right)!}} = \dfrac{{16!61!}}{{64!13!}} = \dfrac{{16.15.14.13!61!}}{{64.63.62.61!13!}} = \dfrac{{16.15.14}}{{64.63.62}} = \dfrac{5}{{372}}\]

Therefore, the chance that the chosen three squares should be in a diagonal line is \[\dfrac{5}{{372}}\].

Hence, option D is correct.

Note- In these type of problems, the chance or the probability can be computed by applying the general formula for probability which is further solved by knowing the possible and total number of outcomes. Here, we are choosing that’s why taking combinations but if we had to arrange then we would have taken permutations.