Aptitude Discussion

Q. |
A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black? |

✖ A. |
18/145 |

✖ B. |
18/29 |

✖ C. |
36/135 |

✔ D. |
36/145 |

**Solution:**

Option(**D**) is correct

The probability that first ball is white:

$= \dfrac{^{12}C_1}{^{30}C_1}$

$= \dfrac{12}{30}$

$=\dfrac{2}{5}$

Since, the ball is not replaced; hence the number of balls left in bag is 29.

Hence, the probability the second ball is black:

$=\dfrac{^{18}C_1}{^{29}C_1}$

$= \dfrac{18}{29}$

Required probability,

$= \left(\dfrac{2}{5}\right) × \left(\dfrac{18}{29}\right)$

$= \dfrac{\textbf{36}}{\textbf{145}}$

**Vaishnavi**

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**Raju**

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Why can't we use combination in this question. i.e., $^{12}c_1 * \dfrac{^{18}c_1}{^{30}c_2}$

But the answer is different from the answer which you get.

Kindly clarify my doubt.

**Ajith Sanjay**

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you can also solve with out using combination,answer will be d same

**Ritu Mehta**

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I am unable to understand that why combination is applied in the ques. 3 and 5..

Combination is used because we have to select the white/black ball from the total number of balls.

Whenever we are required to select the number of objects combination is used..

Combination is used just to count the favourable number of cases so that basic probability formula:

$\textbf{Required Probability}=\dfrac{\textbf{Favourable cases}}{\textbf{Total cases}}$

can be used.

So to count the total number of favourable cases combination is used in these questions.

A container contains hundred eggs. They can be either fresh or rotten. What is sure is the fact that there is at least one fresh egg in that container.

If you are asked to pick two eggs randomly from the container, at least one of them will be rotten.

Can you calculate how many eggs in that container are fresh ?