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  • Solve this problem An urn contains 5 red, 4 black and 3 white marbles. Then the number of ways in which 4 marbles can be drawn from it so that at most 3 of them are red, is :

An urn contains 5 red, 4 black and 3 white marbles. Then the number of ways in which 4 marbles can be drawn from it so that at most 3 of them are red, is :

  • Option 1)

    495

  • Option 2)

    455

  • Option 3)

    460

  • Option 4)

    490

 

Answers (1)

As we learnt in

Theorem of Combination -

Each of the different groups or selection which can be made by taking r things from n things is called a combination.

^{n}c_{r}=\frac{(n)!}{r!(n-r)!}

- wherein

Where 1\leq r\leq n

 

 Total marbles are

5 Red, 4 Black, 3 White marbles.

4 marbles are drawn, atmost 3 of them are red.

Cases are

    3 R    1 other \Rightarrow\ \; ^{5}C_{3}\times^{7}C_{1}=70

    2 R    2 other \Rightarrow\ \; ^{5}C_{2}\times^{7}C_{2}=210

    1 R    3 other  \Rightarrow\ \; ^{5}C_{1}\times^{7}C_{3}=175

Zero R  4 other  \Rightarrow\ \; ^{5}C_{0}\times^{7}C_{4}=\frac{35}{490}

Correct option is 4.

 

 


Option 1)

495

This is an incorrect option.

Option 2)

455

This is an incorrect option.

Option 3)

460

This is an incorrect option.

Option 4)

490

This is the correct option.

Posted by

Sabhrant Ambastha

View full answer

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