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Important for :

1

A shop sells 10 tube lights out of which 3 are defective. Salman buys four tube lights.

A

Following the common explanation, we getProbability that at least two of Salman's tube lights work

= | 29 |

30 |

Hence, option A is correct.

n(S) =

7 of the 10 lamps are not defective.

n(T) =

= | 35 | = | 1 |

210 | 6 |

We need the probability that at least two of his tube lights work.

The event that less than two of his tube lights work,

and the event that at least two of his tube lights work, are exhaustive.

So, we calculate the probability that less than two of his tube lights work and subtract it from 1.

The probability that none of Salman's tube lights work = 0 as there are only 3 defective tube-lights and he buys 4. If K is the probability that only one of Salman's tube lights works,

n(K) =

= | 7 | = | 1 |

210 | 30 |

= 1 – | 1 | = | 29 |

30 | 30 |

2

A shop sells 10 tube lights out of which 3 are defective. Salman buys four tube lights.

E

Following the common explanation, we getProbability that all of Salman's tube lights work

= | 1 |

6 |

Hence, option E is correct.

n(S) =

7 of the 10 lamps are not defective.

n(T) =

= | 35 | = | 1 |

210 | 6 |

We need the probability that at least two of his tube lights work.

The event that less than two of his tube lights work,

and the event that at least two of his tube lights work, are exhaustive.

So, we calculate the probability that less than two of his tube lights work and subtract it from 1.

The probability that none of Salman's tube lights work = 0 as there are only 3 defective tube-lights and he buys 4. If K is the probability that only one of Salman's tube lights works,

n(K) =

= | 7 | = | 1 |

210 | 30 |

= 1 – | 1 | = | 29 |

30 | 30 |

3

From a box containing 8 yellow and 5 white pens, three are drawn one after the other.

4

From a box containing 8 yellow and 5 white pens, three are drawn one after the other.

**Find the probability of all three pens being yellow if the pen drawn is replaced by another yellow colored pen before the next pen is picked.**

B

When the pens drawn are replaced, we can see that the number of pens available for drawing out will be the same for every draw. This means that the probability of a yellow pen appearing in every draw are will be the same.

∴ Reqd probability = |
8 | × | 8 | × | 8 | = | 512 |

13 | 13 | 13 | 2197 |

Hence, option B is correct.

5

A box contains 4 white, 6 green, 2 red and 5 yellow pens.