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Given : 20 out of 144 are defective i.e., no. of defective ball pens = 20 

no. of good ball pens = 144 – 20 = 124 

∴ Total outcomes in drawing a ball pen at random = 144. 

i) Sudha buys it if it is not defective / a good one. No. of outcomes favourable to a good pen = 124. 

∴ Probability of buying it

= \(\frac{No.\,of\,favourable\,outcomes}{Total\,no.\,of\,outcomes}\)

\(=\frac{124}{144}=\frac{31}{36}\)

ii) Sudha will not buy it-if it is a defective pen 

No. of outcomes favourable to a defective pen = 20 

∴ Probability of not buying it

= \(\frac{No.\,of\,favourable\,outcomes}{Total\,no.\,of\,outcomes}\)

\(=\frac{20}{144}=\frac{5}{36}\)

!! (not buying) = 1 – P (buying) 

= 1 – \(\frac{31}{36}=\frac{5}{36}\)

No. of good pens = 144 – 20 = 24

No. of detective pens = 20

Total no. of possible outcomes = 144 {total no pens}

(i) E ⟶ event of buying pen which is good.

No. of favourable outcomes = 124 {124 good pens}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

P(E) = 124/144 = 31/36

(ii) Bar  E⟶ event of not buying a pen which is bad P(E) + P(Bar E) = 1

P(E)+P(Bar E)=1

P(Bar E) = 1 - P(E)

=1−31/36 = 5/36

Outcomes= (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

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

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

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

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

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

(i) P(5 will not come up either time) = 25/36

Explanation:- 5 will not come up either time means In the throw no dice gets a number 5 on them so possible outcomes = (1,1),(1,2),(1,3),(1,4),(1,6)

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

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

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

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

(ii) P(5 will come up at least once) = 11/36

Explanation:- 5 will come up at least once means either of the dice gets a number 5 on them. So possible outcomes are :- (1,5),(2,5),(3,5),(4,5),(6,5) (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

Answer:

(i) Consider the following events.
A = first throw shows 5,
B = second throw shows 5
P(A) = 6/36, P(B) = 6/36 and P(notB) = 5/6
⇒ P(notA) = 1– 6/36 = 30/36 = 5/6
Required probability = 5/6 × 5/6 = 25/36

(ii) Number of events when 5 comes at least once = 11
Probability = 11/36

The correct option is: (C) 3/4

m(S) = 5 + 8 + 10 = 23 

A= Red or Green, n(A)= 5 + 8 =13 

P(A) = \(\frac{n(A)}{n(S)} = \frac{13}{23}\)

(i) Total number of balls 

= 5 + 7 + 8 = 20 

Number of white balls = 5 

Probability of getting white ball is one chance = 5/20 = 1/4

∵ All events are independent 

∴ Required probability = 1/4 × 1/4 × 1/4 × 1/4 = (1/4)⁴ 

(ii) Probability of drawing white ball first time 

= ³C₁ ×1/4 × 1/4 × 1/4 

= 3 ×(1/4)³ 

(iii) P(no ball is white) 

∴ Number of other balls = 7 + 8 = 15 

∴ Probability of drawing one other colour ball = 15/20 = 3/4

∴ Probability of other colour balls drawn successively (none is white) 

= 3/4 × 3/4 × 3/4 × 3/4 

= (3/4)⁴ 

(iv) P(at least 3 white) = P (four white) + P(three white)

= (1/4)⁴ + 3/4³ = 1/4⁴ = 3/4³

= 13/4⁴

The total number of ways of ticking the answers in any one attempt = 2⁴ - 1 = 15. 

The student is taking chance at ticking the correct answer, It is reasonable to assume that in order to derive maximum benefit, the three solutions which he submit must be all different. 

∴ n = total no. of ways = ¹⁵C₃ 

m = the no. of ways in which the correct solution is excluded ¹⁴C₃ 

Hence the required probability = 1 - ¹⁴C₃/¹⁵C₃ = 1 – 4/5 = 1/5 

ALTERNATE SOLUTION: 

The candidate may tick one or more of the alternatives. As each alternative may or may not be chosen, the total numbers of exhaustive possibilities are 2⁴ - 1 = 15. 

Therefore the prob. that the questions are correctly answered by candidate is 1/15.

As such the candidate may be correct on the first, second or third chance. As these events are mutually exclusive, the total probability will be given by 

= 1/15 + 14/15 x 1/14 + 14/15 x 13/14 x 1/13 = 1/15 + 1/15 + 1/15 = 3/15 = 1/5 

Thus the probability that the candidate gets marks in the question is 1/5. 

Since the man is one step away from starting point means that either 

(i) man has taken 6 steps forward aand 5 steps backward 

Or (ii) man has taken 5 steps forward and 6 steps backward. 

Taking movement 1 step forward as success and 1 step backward as failure. 

∴ p = Probability of success = 0.4 and q = probability of failure = 0.6 

∴ required probability = P (X = 6 or X = 5) 

= P (X = 6) + P (X = 5) 

= ¹¹C₆ p⁶ q⁵ + ₁₁C⁵ p⁵q⁶ 

= ¹¹C₅ (p⁶q⁵ + p⁵q⁶) = ¹¹C₅ (p + q) (p⁵ q⁵ ) 

= 11. 10. 9. 8. 7/1. 2. 3. 4. 5 (0.4 + 0.6) (0.4 x 0.6)⁵ 

= 462 x 1 x (0.24)⁵ = 0.37 

Hence the required prob. = 0.37 

Let A₁ be the event that the lot contains 2 defective articles and A₂ the event that the lot contains 3 defective articles. Also let A be the event that the testing procedure ends at the twelfth testing. Then according to the question : 

P (A₁) = 0.4 and P (A₂) = 0.6

Since 0 < P (A₁) <1, 0 < P (A₂) < 1, and P (A₁) + P (A₂) = 1 

∴ The events A₁, A₂ form a partition of the sample space. Hence by the theorem of total probability for compound events, we have

NOTE THIS STEP:

 P(A) = P (A₁) P (AI A₁) + P (A₂) P (AI A₂) . . . . . . . . . . . . . . . . . . . . (1) 

Here P (AI A₁) is the probability of the event the testing procedure ends at the twelfth testing when the lot contains 2 defective articles. This is possible when out of 20 articles; first 11 draws must contain 10 non defective and 1 defective articles and 12^th draw must give a defective article. 

∴ P (AI A₁) = ¹⁸C₁₀ x ²C₁/ ²⁰C₁₁ x 1/9 = 11/190 

Similarly, P (AI A₂) = ¹⁷C₉ x ³C₁/ ²⁰C₁₁ x 1/9 = 11/228 

Now substituting the values of P (AI A₁) and P (AI A₂) in eq. (1), we get 

P(A) = 0.4 x 11/190 + 0.6 x 11/228 = 11/475 + 11/380 = 99/1900

Given: A bag contains one white and one red ball. 

To Find: Write the sample space for the given experiment. 

Explanation: Here, There are 1 white ball , 1 red ball in a bag 

Let us denote White ball with W and Red Ball with W 

According to question, when one ball is drawn, it may be a white or Red. 

So, The sample space of drawing on white ball with replacement 

S_w={(W, W), (W, R)} 

Again, If red ball is drawn , a dice is rolled 

So, The sample space for red ball with dice 

S_R={(R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)} 

Thus, The required sample space for the experiment is: 

S=S_W ∪ S_R 

So, S= {(W, W), (W, R), (R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)} 

Hence, This is the sample space for given experiment

(b) \(\frac{7}{27}.\)

Let S be the sample space of drawing a counter three times and replacing it each time. 

Then, n(S) = 3 × 3 × 3 = 27 

Let A : Event of obtaining a total of 6 in the three draws of counters. 

Then, A = {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1), (2, 2, 2)} 

⇒ n(A) = 7 

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{7}{27}.\)

(b) \(\frac{7}{99}\)

There are (7 + 5) = 12 balls in the bag. 

4 balls can be drawn at random from 12 balls in ¹²C₄ ways. 

∴ n(S) = ¹²C₄ = \(\frac{|\underline{7}}{|\underline3|\underline4}\) = \(\frac{7\times6\times5}{3\times2}\) = 35

∴ Required probability = \(\frac{n(A)}{n(S)}\) = \(\frac{35}{495}\) = \(\frac{7}{99}\).

The helicopter is equally likely to crash anywhere in the region.

Area of the entire region where the helicopter can crash

= (4.5 × 9) km² = 40.5 km²

Area of the lake = (2.5 × 3) km² = 7.5 km²

Therefore, P (helicopter crashed in the lake) =7.5/40.5=75/405=5/27

Let ‘S’ be the sample space. 

So, n(S) = 6² = 36. 

Also, Let ‘A’ and ‘B’ be the events of getting a doublet and a total of 6 respectively. 

∴ A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} 

⇒ n(A) = 6, n(B) = 5 B = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) 

∴ p(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{6}{36}\) , P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{5}{36}\)

A ∩ B = {(3, 3)} ⇒ n(A ∩ B) = 1 

⇒ P(A ∩ B) = \(\frac{n(A∩B)}{n(S)}\) = \(\frac{1}{36}\)

 We know– 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

= \(\frac{6}{36}\) + \(\frac{5}{36}\) − \(\frac{1}{36}\) 

= \(\frac{10}{36}\) 

= \(\frac{5}{18}\)

i. A) \(\frac{1}{8}\)

ii. B) \(\frac{1}{2}\)

iii. A) 1

iv. D) \(\frac{3}{4}\)

(i) Number of times 8 comes = 1

P(8) = 1/8

(ii) Number of Favourable outcomes = 1, 3, 5, 7 = 4

P(an odd number) = 4/8

(iii) Number of Favourable Outcomes = 3, 4, 5, 6, 7, 8 = 6

P(a number greater than 2) = 6/8

(iv) Number of Favourable Outcomes = 1, 2, 3, 4, 5, 6, 7, 8 = 8

P(a number less than 9) = 1

Product of 6 are (1, 6); (2,3); (6, 1); (3,2)

No. of possible out comes = 4

Total number of chances = 6 x 6 = 36

P (Product of 6 ) = 4/36 = 1/9

Total result obtained in throwing a dice.

S = {1,2, 3, 4, 5, 6}

then n(S) = 6

and n{E) = Total result to get digit more than 4 = 2

Thus, required probability

= n(E)/n(s) = 2/6

= 1/3

Total number of days to visit the shop, n(S) = 36 

(i) Number of favorable outcomes, 

n(E) = 6

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

(ii) n(E’) = 1 – n(E) = 1 - \(\frac{1}{6}\) = \(\frac{5}{6}\)

(iii) Number of favorable outcomes, 

n(E) = 5

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{36}\)

The correct option is: (B) 4/5

Explanation:

Shyam and Ekta are visiting a shop from Tuesday to Saturday. 

Total possible ways of visiting the shop by them = 5 x 5 = 25

Possible ways of visiting the shop on same day = 5 

.'. Possible ways of visiting the shop on different days = 25 - 5 = 20 

.'. Probability of visiting the shop on different days = 20/25 = 4/5

Two customers Shyam and Ekta visit a shop on Tuesday to Saturday during a week. The visit not visit made by the each customer is at random. 

All possible outcomes are.

If Tuesday be (T) be Wednesday (W) and so Thursday (Th), Friday (F) and Saturday (S).

The all possible outcomes are :

(T, T), (T, W), (T, Th), (T, F), (T, S)

(W, T), (W, W), (W, Th), (W, F), (W, S)

(Th, T), (Th, W), (Th, Th), (Th, F), (Th, S)

(F, T), (F, W), (F, Th), (F, F), (F, S)

(S, T), (S, W), (S, Tb), (S, F), (S, S)

The total number of all possible outcomes = 25

(i) The favourable outcomes of the visit made by two customers on same day

= (T, T), (W, W), (Th, Th), (F, F), (S, S)

∴ The number of favourable outcomes = 5.

∴ The probability that the two customers visit the shop on same day 

= \(\frac { 5 }{ 25 }\) = \(\frac { 1 }{ 5 }\)

⇒ P(A) = \(\frac { 1 }{ 5 }\)

(ii) The favourable outcomes of visiting two customer on two consecutive days 

= (Shyam, Ekta) = (T, W), (W, Th), (Th, F), (F, S) 

or (Ekta, Shyam) = (W, T), (Th, W), (F, Th), (S, F)

∴ The number of favourable outcomes = 8

∴ The probability that two customers made the visit to the shop on consecutive days = \(\frac { 8 }{ 25 }\)

(iii) The probability ‘not visit’ made by two customers P(\(\overline { A }\)) = 1 – P(A) 

= 1 – \(\frac { 1 }{ 5 }\) = \(\frac { 4 }{ 5 }\)

(i) Odds in favour of X solving a problem are 4 : 3. ∴ The probability of X solving the problem is P(X) = 4/4+3 = 4/7

(ii) Odds against Y solving the problem are 2 : 3. 

∴ The probability of Y solving the problem is P(Y) = 1 – P(Y’)

= 1 – 2/2+3

= 1- 2/5

= 3/5