Mathametics 10 – Statistics and Probability

1) A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that he had wrongly copied down an observation as 50 instead of 40. The correct mean and standard deviation are

(a) 39.9, 6               (b) 36.4, 5               (c) 39.9, 5               (d) none of these  

(d) Let the variable be x. Here n = 100, incorrect A = 40, incorrect σ = 5.1.

From formula A ∑x/n , ∑x = nA

∴ Incorrect ∑x = 100 × 40 = 4000

∴ Correct Σx = incorrect ∑x – 50 + 40 = 4000 – 50 + 40 = 3990

Again, from σ2 + A2 = ∑x2/n ,

We get ∑x2 = n (σ2 + A2)

∴ Incorrect ∑x2 = 100 {(5.1)2 + (40)2} = 100 (26.01 + 1600) = 162601

∴ Correct ∑x2 = incorrect ∑x2 – 502 + 402 = 162601 – 2500 + 1600 = 161701

∴ correct mean A = ∑x/n = 3900/100 = 39.9

Correct (σ2+A2) = ∑x2/n = 161701/100 = 1617.01

∴ Correct σ2 = 1617.01 – (correct A2)

= 1617.01 – (39.9)2

= 1617.01 – 1592.01 = 25

2) In any discrete series (when all values are same), the relationship between MD about mean and SD is

(a) MD = SD               (b) MD ≥ SD               (c) MD < SD               (d) MD ≤ SD  

(b) Let xi/fi; i = 1, 2,…………., n be a frequency distribution.

                                       n

Then SD = √1/N fi (xi – x‾)2

                          n          i=1

MD = 1/N fi |xi – x‾|

                        i=1

Let |xi – x‾| ; i = i,2,….n

                                                            n                             n

Then (SD)2 – (MD)2 = 1/n fi xi2 – [1/n fi zi]2 = σ2 ≥ 0

                                                           i=1                          i=1

⇒ SD ≥ MD

3) A sample of 35 observation has the mean 30 and SD as 4. A second sample of 65 observations from the same population has mean 70 and SD as 3. The SD of the combined sample is

(a) 5.85               (b) 5.58               (c) 3.42               (d) none of these  

(a) Her n1 = 35, x1‾ = 30, σ1 = 4, n2 = 65, x2‾ = 70, σ2 = 3

∴ x12‾ = (35×30+65×70)/(35+65) = 73.5

σ12 = √[35(16+42×25)+65(9+12×25)]/√100

= √34.21

= 5.85

4) The mean deviation of the numbers 3, 4, 5, 6, 7 is

(a) 0              (b) 1.2               (c) 5              (d) 25 

(b) Mean x‾ = (3+4+5+6+7)/5 = 5

N = 5

x                             |x – x‾|

3                                2

4                                1

5                                0

6                                1

7                                2

∑ |x – x‾| = 6

∴ Mean deviation from the mean = ∑ |x – x‾|/N = 6/5 = 1.2

5) If the mean of a set of observations x1, x2,…. xn is x‾ , then the mean of the observations xi + 2i, i = 1, 2, …..n is

(a) x‾+2              (b) x‾+2n               (c) x‾+(n+1)              (d) x‾+n

(c) It is given that x‾ = (x1+x2+…..+xn)/n


Therefore nx‾ = x1+x2+…..+xn


Let y‾ be the mean of observations xi +2n, i = 1, 2, ….., n


Then y‾ = (x1+2.1)+(x2+2.2)+(x3+2.3)+……+(xn+2n)

        n

= [ xi+2(1+2+3+…+n)]/n

       i=1  

               n

= 1/n xi + [2(n)(n+1)]/2n = x‾ +(n+1)

             i=1

6) If a is the GM of the product of r sets of observations with geometric means G1,G2, …Gr respectively, then G is equal t

(a) G1/G2              (b) logG1 – logG2               (c) logG1/logG2              (d) log(G1,G2) 

(a) Let x1, x2,………xn and y1, y2,……., yn be two series of observations with geometric means G1 and G2 respectively.

Then, G1 = (x1, x2 ………xn)1/n

and G2 = (y1, y2….. yn)1/n

Since G is the geometric mean of the ratios of the corresponding observations, therefore

G = [(x1/y1) , (x2/y2) ,…, (xn/yn)]1/n

= [(x1,x2,…xn)1/n]/[(y1,y2,…,yn)1/n] = G1/G2

7) One number is selected at random from first two hundred positive integers. The probability that it is divisible by 6 or 8, is equal to

(a) 29/100             (b) 33/100               (c) 43/100              (d) 1/4 

(a) Total number of integers that are divisible by six = (200/6)+(200/36) = 38

Total number of integers that are divisible by eight = (200/8)+(200/64) = 28

Total number of integers that are divisible by six as well as by eight i.e. by 24

= 200/24 = 8

Thus total number of integers that are divisible by 6 or 8

= 38 + 28 – 8 = 58

Thus, required probability = 58/200 = 29/100

8) Two fair and ordinary dice are rolled simultaneously. The probability of getting the sum of outcomes of the dice as a multiple of 4, is equal to

(a) 5/9            (b) 1/4               (c) 1/9              (d) 1/3 

(b) In the case sum can be 4 or 8 or 12.

Total ways of getting the sum as 4

= 3 (namely (2, 2), (1, 3), (3, 1).

Total ways of getting the sum as 8

= 5 (namely (2, 6), (6, 2), (3, 5), (5, 3), (4, 4))

Total ways of getting the sum as 12

= 1 (namely (6, 6)

Thus total ways of getting the sum as multiple of four

= 3 + 5 + 1 = 9

Thus, required probability = 9/36 = 1/4

9) A five digit number (having all different digits) is formed using the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. The probability that the formed number either begins or ends with an odd digit, is equal to

(a) 5/6            (b) 1/6               (c) 1/3              (d) 2/3 

(a) Total formed numbers that begin with a odd digit

= 5C1 × 8P4 = 5(8) (7) (6) (5)

Total formed numbers that end with a odd digit

= 5C1 × 8P4 = (8) (7) (6) (5)

Total formed number that begin with an odd digit and also end with an odd digit

= 5C2 × 2! × 7P3 = 5 × (4) (7) (6) (5)

Thus total formed numbers that begin with an odd digit or end with an odd digits is equal to 5 × 7 × 6 × 60

Total formed numbers = 9P5 = 9 × 8 × 7 × 6 × 5

Thus, required probability = 5/8

10) The probabilities of solving a problem by students A, B and C independently are 1/2 , 1/3 , and 1/4 respectively. If they start solving the given problem. Independently, then the probability that atleast two of them will solve the problem successfully, is equal to

(a) 5/24            (b) 9/24               (c) 7/24              (d) 11/24 

(c) Probability that A and B can solve but ‘C’ is unable to solve

= (1/2)×(1/3)×(3/4) = 1/8

Probability that A and C can solve but ‘B’ is unable to solve

= (1/2)×(2/3)×(1/4) = 1/12

Probability that B and C can solve but ‘A’ is unable to solve

= (1/2)×(1/3)×(1/4) = 1/24

Probability that all of them can solve the problem

= (1/2)×(1/3)×(1/4) = 1/24

Thus, required probability that atleast two of them can solve the problem

= 7/24

11) A bag contains 4 red, 5 white and 6 black balls. Three balls are selected from this bag simultaneously. The probability that one of the colour will be missing in the selected balls, is equal to

(a) 301/455            (b) 366/455               (c) 261/455              (d) None of these

(a) (i) Selected balls can be W W B or W B B

(when red colour is missing).

Corresponding probability = (5C2×6C1+5C1×5C2)/15C3 = 135/15C3

(ii) Selected balls can be R R W or R W W

(when black colour is missing).

Corresponding probability = (4C2×5C1+4C1×5C2)/15C3 = 70/15C3

(iii) Selected balls can be R B B, R R B

(when white colour is missing).

Corresponding probability = (4C1×6C2+4C2×6C1)/15C3 = 96/15C3

Thus, required probability = 301/455

12) In a certain city 40% of the people have brown hair, 25% have brown eyes and 15% have brown eyes as well as brown hair. If a person selected at random has brown hair, the probability that the also has brown eyes, is equal to

(a) 2/6            (b) 1/4               (c) 1/2              (d) 3/8

(d) E1 : Person has brown hair.

E2 : Person has brown hair.

P(E1) = 40/100 = 2/5

P(E2) = 25/100 = 1/4

P(E1∩E2) = 15/100 = 3/20

P(E2/ E1) = P(E1∩E2)/P(E1)

= (3/20)/(2/5) = 3/8

13) A pair of fair and ordinary dice is rolled simultaneously. It is found that they show different outcomes. The probability that the sum of the outcomes will be either 6 or 10, is equal to

(a) 1/5            (b) 1/6               (c) 1/3              (d) 2/3

(a) E1 : Event that dice show different outcomes.

E2 : Outcome is either 6 or 10.

P(E1) = 5/6 , P(E1∩E2) = 1/6

As 6 can occur in four ways namely

(2 , 4), (4 , 2) (1 , 5), (5 , 1)

And ten can occur in 2 ways namely (6 , 4) and (4 , 6).

Thus, required probability,

P(E1/E2) = P(E1∩E2)/P(E1) = (1/6)/(5/6) = 1/5

14) Two fair dice are rolled simultaneously. It is found that one of them shows odd prime numbers. The probability that remaining dice also shows an odd prime number, is equal to

(a) 1/5            (b) 2/5               (c) 3/5              (d) None of these

(a) 3 and 5 are the only odd prime numbers, among the possible outcomes.

The following are the outcomes when one of themshow odd prime numbers,

(3, 1), (1, 3) (3, 2), (2, 3) (3, 3), (3, 4) (4, 3), (3, 5) (5, 3), (3, 6), (6, 3), (5, 1), (1, 5), (5, 2), (2, 5), (5, 4), (4, 5), (5, 5), (5, 6), (6, 5).

Out of these 20 equally likely outcomes exactly 4 favour the presence of odd prime numbers on both dice.

Thus, required probability = 4/20 = 1/5

15) A bag contains four tickets marked with numbers 112, 121, 211, 222. One ticket is drawn at random from the bag. Let Ei (i = 1, 2, 3) denote the event that ith digit on the selected ticket is 2, then which of the following is not correct?

(a) E1 and E2 are independent                          (b) E2 and E3 are independent               

(c) E3 and E1 are independent                          (d) E1 , E2 , E3 are independent

(d) P(E1) = 2/4 = 1/2,

P(E2) = 2/4 = 1/2,

P(E3) = 2/4 = 1/2

P(E1∩E2) = 1/4 = P(E1).(E2)

⇒ ‘E1‘ and ‘E2‘ are independent.

P(E1∩E3) = 1/4 = P(E1).(E3)

⇒ ‘E1‘ and ‘E3‘ are independent.

P(E2∩E3) = 1/4 = P(E2).(E3)

⇒ ‘E2‘ and ‘E3‘ are independent.

P(E1∩E2∩E3) = 1/4 = P(E1).(E2).(E3)

⇒ ‘E1‘ , ‘E2‘ and ‘E3‘ are not independent.

16) Two numbers and are chosen at random (without replacement) from the set {1, 2, 3, …., 5n}. The probability that n14 – n24 is divisible by 5, is equal to

(a) (n-1)/(5n-1)           (b) 4(4n-1)/5(5n-1)               (c) (17n-5)/5(5n-1)              (d) 8n/5(5n-1)

(c) n14 – n24 = (n12+n22)(n1+n2)(n1-n2)

We can segregate the numbers as multiple of five (5λ), 5λ+ 1, 5λ+ 2, 5λ+ 3, 5λ+ 4, as follows:

R1 → 1     6……..5n-4 → 5λ+1

R2 → 2     7……..5n-3 → 5λ+2

R3 → 3     8……..5n-2 → 5λ+3

R4 → 4     9……..5n-1 → 5λ+4

R5 → 5    10……..5n- → 5λ

We can select either both the number from
R5 , or any two numbers from the first four rows.

(As square of any number that is not a multiple of three will be in the form of 5λ ’ + 1 or 5λ ’ – 1)

Thus, required probability

= (nC2+4nC2)/5nC2

= [n(n-1)+4n(4n-1)]/5n(5n-1)

= (17n-5)/5(5n-1)

17) A man alternatively tosses a fair coin and rolls a fair ordinary dice. He starts with the coin. The probability that he gets a tail on the coin before getting 5 or 6 on the dice, is equal to

(a) 3/4           (b) 1/2               (c) 1/3              (d) 2/3

(a) Probability of getting 5 or 6 on a specific roll of dice = 2/6 = 1/3


and, probability of getting a tail = 1/2


The desired outcome can happen, in general, on (2r + 1 )th trial.


That means first 2r trials should result neither in tail nor in 5 or 6, and (2r + 1 )th trial must result in tail.


It the corresponding probability is pr , then


pr = (1/2)r × (2/3)r ×(1/2)


= (1/2)×(1/3)r


Thus, required probability =

 ∞                ∞

pr = 1/2 (1/3)r = 3/4 

r=0              r=o

18) A committee consists of 9 experts taken from three institutions A, B and C, of which 2 are from A, 3 from B and 4 from C. If three experts resign from the committee, then the probability of exactly two of the resigned experts being from the same institution, is equal to

(a) 4/7           (b) 25/84               (c) 55/84              (d) 37/84

(c) These are three mutually exclusive cases:

(i) 2 are from A and another is from either from B or C.

Corresponding probability = 2C2 . (3C1+4C1)/9C3 = 7/84

(ii) 2 are from B and another is either from A or C.

Corresponding probability = 3C2 . (2C1+4C1)/9C3 = 18/84

(iii) 2 are from C and another is either from A or B.

Corresponding probability = 4C2 . (2C1+3C1)/9C3 = 30/84

Thus, required probability = (7+18+30)/84 = 55/84

19) The sum of two natural numbers n1 and n2 is known to be equal to 100. The probability that their product being greater than 1600, is equal to

(a) 20/33           (b) 58/99               (c) 13/33              (d) 59/99

(d) Total number of ways in which n1+n2 = 100 is equal to 99.

Now,
n1n2 > 1600

n1(100-n1) > 1600

n12 – 100n+1600 < 0

(n1-80)(n1-20) < 0

20 < n1 < 80

21 ≤ n1 ≤ 79 

Thus, number of favorable ways= 79 – 21 + 1 = 59

Hence, required probability = 59/99

20) A bag contains ‘W’ white balls and ‘R’ red balls. Two players P1 and P2 alternately draw a ball from the bag, replacing the ball each time after the draw, till one of them draws a white ball and wins the game. ‘P1‘ Beings the game. The probability of ‘P2‘ being the winner , is equal to

(a) W2/(W+2R)R           (b) R/(2W+R)               (c) R2(2W+R)W             (d) R/(W+2R)

(d) Probability of drawing a white ball at any draw = W/(W+R)


Now, P2 can be the winner in general on 2r th, r ≥ 1, draw. That means in the first (2r – 1) draws no player draws a 


white ball and 2r th draw results in a white ball for P2


If the corresponding probability is pr , then


pr = [R/(W+R)]r × [R/(W+R)]r-1 ×[W/(W+R)


[R/(W+R)]2r × (W/R)


Hence, required probability

    ∞                   ∞

= pr = W/R [R/(W+R)]2r

   r=1                 r=1


= R/(W+2R)

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