**From a bag containing 9 distinct white and 9 distinct black, 9 balls are drawn at random one by one, the drawn balls being replaced each time. The probability that at least four balls of each colour is in the draw, is**

__Question:__**Options:**

A. | A little less than 12 | |

B. | a little greater than 12 | |

C. | 12 | |

D. | 13 |

**Answer: Option A**

: A

Out of 9 distinct black and 9 distinct white balls, probability of drawing a white ball =12

and drawing black ball is also 12. for at least 4 of each colour in 9 draws with replacement, there are two cases,

Case(i)

P(getting 5 white, 4 black) =9C5(12)9

Cases (ii)

P(getting 4 white, 5 black)=9C4(12)9

These are exclusive so

P(atleast 4 of each colour) =9C5(12)9+9C4(12)9=63128 which is little less than 12

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## More Questions on This Topic :

**In a multiple choice question there are four alternative answers of which one or more than one is or are correct. A candidate will get marks on the question only if he ticks all correct answers. The candidate decides to tick answers at random. If he is allowed up to three chances to answer the question, the probability that he will get marks on it is given by**

__Question 1.__- 1−(1415)3
- (115)3
- 15
- 1415

**Answer: Option C**

: C

The total number of ways of answering one or more alternatives of 4 is 4C1+4C2+4C3+4C4=15,

Out of these 15 combinations, only one combination is correct. The probability of answeringthe alternaive correctly at the first trial is 115, that at the second trial is (1415.114)=115, and that at the third trail is 1415.1314.113=115,

Therefore the probability that the andidate will get marks on the question if he allowed upto three chances, is 115+115+115=15.

**The odds in favour of A solving a problem are 3 to 4 and the odds against B solving the same problem are 5 to 7. If they both try the problem, the probability that the problem is solved is:**

__Question 2.__- 4184
- 1621
- 521
- 14

**Answer: Option B**

: B

P(A)=37,P(B)=712P(¯A)=47,P(¯B)=512P(A∪B)=1−P(¯¯¯¯¯¯¯¯¯¯¯¯¯¯A∪B)=1−P(¯A∩¯B)∴P(A∪B)=1−47×512=1621

**Urn A contains 6 red , 4 white balls and urn B contains 4 red and 6 white balls. One ball is drawn from the urn A and placed in the urn B. Then one ball is drawn at random from urn B and placed in the urn A. if one ball is now drawn from the urn A, then the probability that it is found to be red is**

__Question 3.__- 3255
- 3040
- 3250
- 34

**Answer: Option A**

: A

case(i): red ball from A to B, red ball from B to A ,then red ball from A

P1=610×511×610

Case(ii): red ball from A to B, white ball from B to A, red ball from A

P2=610×611×510

Case(iii): white ball from A to B, red ball from B to A, red ball from A

P3=410×411×710

Case(iv): white ball from A to B, white ball from B to A, red ball from A

P4=410×711×610

Therefore required probability =P1+P2+P3+P4=6401100=3255

**Three six faced fair dice are rolled together. The probability that the sum of the numbers appearing on the dice is 8, is**

__Question 4.__- 772
- 754
- 427
- 764

**Answer: Option A**

: A

Getting sum 8 when three dice are rolled = E

The possibilities are (1,1,6) , (2,2,4), (3,3,2), (1,2,5), (1,3,4)

The number of favorable cases for E = 3 + 3 + 3 + 6 + 6 = 21

The total number of cases = 63= 216

Required probability P(E) =21216=772.

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