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6 months ago

Probability Probability

37. Given P(A) = $\frac{3}{5}$ and P(B) = $\frac{1}{5}$ . Find P(A or B), if A and B are mutually exclusive events.

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alok kumar singh

Contributor-Level 10

37.

Given P (A) = $\frac{3}{5}$

P (B) = $\frac{1}{5}$

As A and B are mutually exclusive events,

P (A∩B) = 0

Hence, P (A∪B) = P (A) + P (B) – P (A∩B)

$= \frac{3}{5} + \frac{1}{5} - 0$

$= \frac{4}{5}$

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New answer posted

6 months ago

Probability Probability

36. Fill in the blanks in following table:

P(A) P(B) P(A ∩ B) P(A ∪ B)

(i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$ . . .

(ii) 0.35 . . . 0.25 0.6

(iii) 0.5 0.35 . . . 0.7

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alok kumar singh

Contributor-Level 10

36. (i) Given P (A) = $\frac{1}{3}$

P (B) = $\frac{1}{5}$

P (A∩B) = $\frac{1}{15}$

So, P (A∪B) = P (A) + P (B) – P (A∩B)

$= \frac{1}{3} + \frac{1}{5} - \frac{1}{15}$

$= \frac{5 + 3 - 1}{15}$

P (A∪B) = $\frac{7}{15}$

(ii) Given P (A) = 0.35

P (B) =?

P (A∩B) = 0.25

P (A∪B) = 0.6

So, P (A∪B) = P (A) + P (B) – P (A∩B)

0.6 = 0.35 + P (B) – 0.25

P (B) = 0.6 – 0.35 + 0.25

P (B) = 0.5

(iii) Given P (A) = 0.5

P (B) = 0.35

P (A∩B) =?

P (A∪B) = 0.7

So, P (A∪B) = P (A) + P (B) – P (A∩B)

0.7 = 0.5 + 0.35 – P (A∩B)

P (A∩B) = 0.5 + 0.35 – 0.7

P (A∩B) = 0.15

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New answer posted

6 months ago

Probability Probability

35. Check whether the following probabilities P(A) and P(B) are consistently defined

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6

(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

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alok kumar singh

Contributor-Level 10

35. Given P (A) = 0.5

P (B) = 0.7

And P (A∩B) = 0.6

As P (A∩B) > P (A) which is not possible.

The given probabilities are not consistently defined.

(ii) Given, P (A) = 0.5

P (B) = 0.4

And P (A∪B) = 0.8

So, P (A∪B) = P (A) + P (B) – P (A∩B)

0.8 = 0.5 + 0.4 – P (A∩B)

P (A∩B) = 0.5 + 0.4 – 0.8

P (A∩B) = 0.1

Hence, P (A∩B) < P (A) and P (AB) < P (B)

The given probabilities are consistently defined.

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New answer posted

6 months ago

Probability Probability

34. In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint order of the numbers is not important.]

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alok kumar singh

Contributor-Level 10

34.

. Since 6 numbers are to be choosen as fixed from a set a given 20 number, the sample space is

$n () = 20 C_{6} = \frac{20!}{6! (20 - 6)!} = \frac{20 * 19 * 18 * 17 * 16 * 15 * 14!}{2 * 3 * 4 * 5 * 6 * 14!}$

$= \frac{38760}{1} * \frac{14!}{14!} = 38760$

Let A: person wins the prize.

In order to win the prize the 6 number has to be correct i.e. all 6 of the number are to be choosen from fixed 6 numbers we have,

$n (S) =^{6}_{C6} = 1$

? P (A) = $\frac{1}{38760}$ .

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New answer posted

6 months ago

Probability Probability

33. A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that letter is (i) A vowel (ii) A consonan

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alok kumar singh

Contributor-Level 10

33. The sample space of word is

S = {A, S, A, S, I, N, A, T, I, O, N}

So, n (S) = 13.

(i) Let A: word is a vowel

A = {A, I, A, I, O}

So, n (A) = 6

? P (A) = $\frac{6}{13}$

(ii) Let B: Word is a consonant

B = {S, N, T, N}

So, n (B) = 7

? P (B) = $\frac{7}{13}$ .

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New answer posted

6 months ago

Probability Probability

32. If $\frac{2}{11}$ is the probability of an event, what is the probability of the event 'not A'.

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alok kumar singh

Contributor-Level 10

32. Let A be the event

Given that, P (A) = $\frac{2}{11}$

So, P (not A) = P (S) – P (A) = $1 - \frac{2}{11} = \frac{11 - 2}{11} = \frac{9}{11}$

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New answer posted

6 months ago

Probability Probability

31. Three coins are tossed once. Find the probability of getting

(i) 3 heads

(ii) 2 heads

(iii) Atleast 2 heads

(iv) Atmost 2 heads

(v) No head

(vi) 3 tails

(vii) Exactly two tails

(viii) No tail

(ix) Atmost two tails

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alok kumar singh

Contributor-Level 10

31. When three coins are tosses we have the sample space,

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

So, n (S) = 8

(i) Let A: 3 heads occurs.

A = {HHH}

So, n (A) = 1

? P (A) = $\frac{1}{8}$

(ii) Let B: 2 heads occurs

B = {HHT, HTH, THH}

So, n (B) = 3

? P (B) = $\frac{3}{8}$

(iii) Let C: at least 2 heads occurs i.e. 2 heads or more

C = {HHT, HTH, THH, HHH}

So, n (C) = 4

? P (C) = $\frac{4}{8} = \frac{1}{2}$

(iv) Let D: at most 2 heads occurs i.e. 2 heads or less

D = {TTT, HTT, THT, TTH, HHT, HTH, THH}

So, n (D) = 7

? P (D) = $\frac{7}{8}$

(v) Let E: no head occurs

E = {TTT}

So, n (E) = 1

? P (E) = $\frac{1}{8}$

(vi) Let F: 3 tails occurs

F = {TTT}

So, n (F) = 1

? P (F) = $\frac{1}{8}$

(vii) Let G: exactly two tail

...more

31. When three coins are tosses we have the sample space,

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

So, n (S) = 8

(i) Let A: 3 heads occurs.

A = {HHH}

So, n (A) = 1

? P (A) = $\frac{1}{8}$

(ii) Let B: 2 heads occurs

B = {HHT, HTH, THH}

So, n (B) = 3

? P (B) = $\frac{3}{8}$

(iii) Let C: at least 2 heads occurs i.e. 2 heads or more

C = {HHT, HTH, THH, HHH}

So, n (C) = 4

? P (C) = $\frac{4}{8} = \frac{1}{2}$

(iv) Let D: at most 2 heads occurs i.e. 2 heads or less

D = {TTT, HTT, THT, TTH, HHT, HTH, THH}

So, n (D) = 7

? P (D) = $\frac{7}{8}$

(v) Let E: no head occurs

E = {TTT}

So, n (E) = 1

? P (E) = $\frac{1}{8}$

(vi) Let F: 3 tails occurs

F = {TTT}

So, n (F) = 1

? P (F) = $\frac{1}{8}$

(vii) Let G: exactly two tails occurs

G = {TTH, THT, HTT}

So, n (G) = 3

? P (G) = $\frac{3}{8}$ .

(viii) Let H: no tail occurs

H = {HHH}

So, n (H) = 1

? P (H) = $\frac{1}{8}$ .

(ix) Let I: atmost two tails i.e. two tails or less

I = {HHH, THH, HTH, HHT, HTT, THT, TTH}

So, n (I) = 7

? P (I) = $\frac{7}{8}$

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New answer posted

6 months ago

Probability Probability

30. A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up.

From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

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alok kumar singh

Contributor-Level 10

30. When a coin is tossed four times we have the sample space,

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, THHT, TTHH, THTH, HTHT, HTTH, TTTH, TTHT, THTT, HTTT, TTTT}

So, n (S) = 16.

Case I: When the outcome is all head, the amount is 1 + 1 + 1 + 1 =? 4 gain

Case II: When the outcome is 3 head and one tail, the amount is

1 + 1 + 1 – 1.50 = 3 – 1.50 =? 1.50 gain

Case III: When the outcome is 2 head and 2 tail, the amount is

1 + 1 – 1.50 – 1.50 = 2 – 3 =? 1 lose.

Case IV: When the outcome is 1 head and 3 tail, the amount is

1 – 1.50 – 1.50 – 1.50 = 1 – 4.50 =? 3.50 lose.

Case V: When the outcome is all tail, the amount is

–1.5

...more

30. When a coin is tossed four times we have the sample space,

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, THHT, TTHH, THTH, HTHT, HTTH, TTTH, TTHT, THTT, HTTT, TTTT}

So, n (S) = 16.

Case I: When the outcome is all head, the amount is 1 + 1 + 1 + 1 =? 4 gain

Case II: When the outcome is 3 head and one tail, the amount is

1 + 1 + 1 – 1.50 = 3 – 1.50 =? 1.50 gain

Case III: When the outcome is 2 head and 2 tail, the amount is

1 + 1 – 1.50 – 1.50 = 2 – 3 =? 1 lose.

Case IV: When the outcome is 1 head and 3 tail, the amount is

1 – 1.50 – 1.50 – 1.50 = 1 – 4.50 =? 3.50 lose.

Case V: When the outcome is all tail, the amount is

–1.50 – 1.50 – 1.50 – 1.50 = –6 =? 6 lose.

Let A be event such that person wins ?4. Then,

A = {HHHH}

So, n (A) = 1

? P (A) = $\frac{1}{16}$

Let B: person wins ?1.5. Then,

B = {HHHT, HHTH, HTHH, THHH}

So, n (B) = 4

? P (B) = $\frac{4}{16} = \frac{1}{4}$

Let C: person loses ?1. Then,

C = {HHTT, THHT, TTHH, THTH, HTHT, HTTH}

So, n (C) = 6

? P (C) = $\frac{6}{16} = \frac{3}{8}$

Let D: person loses ?3.50. Then,

D = {HTTT, THTT, TTHT, TTTH}

So, n (D) = 4

? P (D) = $\frac{4}{16} = \frac{1}{4}$

Let E: person loses ?6. Then,

E = {TTTT}

So, n (E) = 1

? P (E) = $\frac{1}{16}$ .

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New answer posted

6 months ago

Probability Probability

29. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

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alok kumar singh

Contributor-Level 10

29. Number of women in the city council n (A) = 6

As there are four men and six women the total number of person in the sample space is 4 + 6 = 10.

So, n (S) = 10

P (A) = $\frac{n (A)}{n (S)} = \frac{6}{10} = \frac{3}{5}$

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New answer posted

6 months ago

Probability Probability

28. A fair coin with 1 marked on one face and 6 on the other and a fair die are bothtossed. find the probability that the sum of numbers that turn up is (i) 3 (ii) 12

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alok kumar singh

Contributor-Level 10

28. The sample space of the experiment is

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

So, n (S) = 12.

(i) Let E be event such that sum of numbers that turn up is 3. Then,

E = { (1, 2)}

So, n (E) = 1

P (E) = $\frac{n (E)}{n (S)} = \frac{1}{12}$ .

(ii) Let F be event such that sum of number than turn up is 12. Then,

F = { (6, 6)}

So, n (F) = 1

P (F) = $\frac{n (F)}{n (S)} = \frac{1}{12}$ .

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