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Maths Probability

40. A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16.Determine

(i) P(not A), (ii) P(not B) and (iii) P(A or B)

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alok kumar singh

Contributor-Level 10

40. Given P (A) = 0.42

P (B) = 0.48

P (A∩B) = 0.16

(i) P (not A) = P (A') = 1 – P (A) = 1 – 0.42 = 0.58

(ii) P (not B) = P (B') = 1 – P (B) = 1 – 0.48 = 0.52

(iii) P (A or B) = P (A∪B) = P (A) + P (B) – P (A∩B)

= 0.42 + 0.48 – 0.16

= 0.74

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New answer posted

a year ago

Maths Probability

39. Events E and F are such that P(not E or not F) = 0.25, State whether E and F aremutually exclusive.

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alok kumar singh

Contributor-Level 10

39. Given, P (not E or not F) = 0.25

P (E'∪F') = 0.25

P (E∩F)' = 0.25

1 – P (E∩F) = 0.25

P (E∩F) = 1 – 0.25

P (E∩F) = 0.75≠ 0

Hence, E and F are not mutually exclusive events.

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New answer posted

a year ago

Maths Probability

38. If E and F are events such that P(E) = $\frac{1}{4}$ , P(F) = $\frac{1}{2}$ and P(E and F) = $\frac{1}{8}$ , find

(i) P(E or F), (ii) P(not E and not F).

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

38. Given, P (E) = $\frac{1}{4}$

P (F) = $\frac{1}{2}$

P (E and F) = P (E∩F) = $\frac{1}{8}$

(i) P (E or F) = P (E∪F) = P (E) + P (F) – P (E∩F)

$= \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$

$= \frac{2 + 4 - 1}{8}$

$= \frac{5}{8}$

(ii) P (not E and not F) = P (E'∩F') = P (E∪F)' = 1 – P (E∪F)

$= 1 - \frac{5}{8} = \frac{8 - 5}{8}$

$= \frac{3}{8}$

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New answer posted

a year ago

Maths 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.

0 Follower 3 Views

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

a year ago

Maths 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

0 Follower 8 Views

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

a year ago

Maths 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

0 Follower 4 Views

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

a year ago

Maths 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.]

0 Follower 6 Views

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

a year ago

Maths Probability

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

0 Follower 3 Views

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

a year ago

Maths Probability

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

0 Follower 3 Views

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

a year ago

Maths 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

0 Follower 4 Views

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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