**DATA ANALYSIS AND ITERPRETATION**

**Problems based on Combinations & Probability !**

**Problem –**An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles.

Two marbles are picked randomly. What is the probability that both are red.

a. 1/6

b. 1/3

c. 2/15

d. 2/5

e. None

Answer – First one ball can be picked out of six and next time one out of five. So total ways of pick 2 red =

^{6}C_{2}= 6! / ( 4! X 2! ) = 15Total number of picking any two balls =

^{15}C_{2}= 105P(R) =

^{6}C_{2}/^{15}C_{2}= 15 / 105 = 1/7

**Problem –**If three marbles are picked randomly what is the probability that two are blue and one is yellow.

a. 3/91

b. 1/5

c. 18/455

d. 7/15

e. None

**Answer**– Picking three balls, total number of exhaustive cases =

^{15}C

_{3}

Picking three balls, in required fashion =

^{4}C_{2 }x^{3}C_{1}Required probability =

^{4}C_{2 }x^{3}C_{1 }/^{15}C_{3}= 6 x 3 / 455 = 18 / 455

**Problem –**If four marbles are picked at random, then what is the probability that atleast one is blue.

a. 4/15

b. 69/91

c. 11/15

d. 22/91

e. None

**Answer**– Picking 4, total No of exhaustive cases =

^{15}C

_{4}

No of cases when we will have none blue = picking 4 balls from remaining 11 other colour balls =

^{11}C_{4}If we subtract this from total number of cases we get cases where atleast one ball will be blue

=

^{15}C_{4 }-^{11}C_{4 }= 1365 – 330 = 1035Probability = 1035 / 1365 = 207 /273 =69/91

Now you must have understood how to deal with such types.

**Problem**– Study the table carefully and answer the questions that follow.

Number of People staying in Five different localities and the % breakup of men, women and chidren | ||||

Locality | Total No of People | Percentage | ||

Men | Women | Children | ||

F | 5640 | 55 | 35 | 10 |

G | 4850 | 34 | 44 | 22 |

H | 5200 | 48 | 39 | 13 |

I | 6020 | 65 | 25 | 10 |

J | 4900 | 42 | 41 | 17 |

**Question –**What is the total number of men and children staying together in locality I ?

a. 4115

b. 4515

c. 4551

d. 4155

Answer – Total number of men and children staying together in locality I = ( 65% + 10% )of 6020

=(0.65+0.10) x 6020 = 0.75 x 6020 = 4515

**Quesiton –**The number of women staying in which locality is highest ?

a. H

b. J

c. F

d. G

Answer - Number of women staying in each of these 4 localities are

In H = 39% of 5200 = 0.39 x 5200= 2028

In J = 41% of 4200 = 0.41 x 4200 = 2009

In F = 35% of 5640 = 0.35 x 5640 = 1974

In G = 44% of 4850 = 0.44 x 4850 = 2015

Clearly in H locality there are highest number of women.

**Question**– What is the total number of children staying together in ,locality H & I together ?

a. 1287

b. 1278

c. 1827

d. 1728

Answer - Number of children in H = 13% of 5200 = 0.13 x 5200 = 676

Number of Children in I = 10% of 6020 = 0.10 x 6020 = 602

Total number of children staying together in I &H locality together = 676 + 602 = 1278

**Question -**The total number of people staying in locality J forms what percentage of total number of people staying in locality F ?

a. 81

b. 72

c. 78

d. 93

e. 87

Answer - People in locality J divded by people in locality F = 4900 / 5640 = 0.8687 ≈ 0.87 or 87 %

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