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Let A be the event that Dimpu gets scholarship and B the event that Pintu gets scholarship. It is given that P (A) = 0.9, P (B) = 0.8. 

The probability that none of them gets the scholarship = (1 − 0.9) (1 − 0.8) = 0.1 × 0.2 = 0.02 

∴ The probability that at least one of them gets the scholarship = 1 − 0.02 = 0.98.

Let E = {part A is defective}, F = {part B is defective}.

Then, P (E) = \(\frac{9}{100}\), P(F) = \(\frac{5}{100}\)

∴ P(\(\bar{E}\)) = 1 - \(\frac{9}{100}\) = \(\frac{91}{100}\) and   P(\(\bar{F}\)) = 1 - \(\frac{5}{100}\) = \(\frac{95}{100}\)

Since E and F are independent, therefore, \(\bar{E}\) and \(\bar{F}\) and are also independent.

Now P (none is defective) = P(\(\bar{E}\)) . P(\(\bar{F}\)) = \(\frac{91}{100}\) x \(\frac{95}{100}\)

= 0.91 x 0.95 = 0.8645 = 0.86 to 2 d.p.

(c) 0.8645

Required probability = P(X not defective and Y not defective) 

= P(\(\bar{X}\)) x P(\(\bar{Y}\))

= (1 – P(X)) (1 – P(Y))

= \(\bigg(1-\frac{9}{100}\bigg)\)\(\bigg(1-\frac{5}{100}\bigg)\)

= \(\frac{91}{100}\) x \(\frac{95}{100}\) = \(\frac{8645}{10000}\) = 0.8645.

Let E₁ be the event of selecting Bag 1 and E₂ be the event of selecting Bag 2.

Also, let E₃ be the event that black ball is selected

Now,

P(E₁) = 2/6 = 1/3 and P(E₂) = 1 – 1/3 = 2/3

P(E₃/E₁) = 3/7 and P(E₃/E₂) = 4/7

So,

P(E₃) = P(E₁). P(E₃/E₁) + P(E₂). P(E₃/E₂)

= 1/3. 3/7 + 2/3. 4/7 = (3 + 8)/ 21 = 11/21

Therefore, the required probability is 11/21.

Given: P(A) = 0.4, P(A or B) = 0.6 and P(B) = 0.5

We know that, P(A or B) = P(A) + P(B) – P(A and B)

By substituting values in the above formula, we get

0.6 = 0.4 + 0.5 – P(A and B)

0.6 = 0.9 – P(A and B)

P(A and B) = 0.9 – 0.6

P(A and B) = 0.3

P(A and B) = 0.3

Given: P(A or B) = 0.7, P(A and B) = 0.3 and P(bar A) = 0.4

We know, P(A) = 1 – P(bar A)

= 1 – 0.4 = 0.6

So, P(A) = 0.6

Again, P(A or B) = P(A) + P(B) – P(A and B)

By substituting values in the above formula, we get

0.7 = 0.6 + P(B) – 0.3

0.7 = 0.3 + P(B)

0.7 – 0.3 = P(B)

Or P(B) = 0.4

Correct option  (c)  3/5

Explanation:

For real roots, D ≥ 0

⇒  p²– p – 2 ≥ 0

⇒  p ≤ –1 p ≥ 2

∴ p  ∈ [2,5] but p ∈ [0,5]

∴ required probability  = 5 -2/5 - 0 = 3/5

No. of all outcomes=(30green + 30white)balls

=60balls

No. of event= no.of white balls

=30

So, probability of drawing white balls=n(E)/n(S)

=30/60

=1/2=0.5

Total cards = 5

(i) P(Queen) = 1/5

(ii) P(Ace ) = 1/4   (Since, Queen was kept aside)

Correct option is: D)\(\frac{7}{8}\)

When three coins are tossed, then possible outcomes are

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

\(\therefore\) Total possible outcomes = n(s) = 8

Outcomes which favours the event that at least are head appears are 

{TTH, THT, TTH, THH, HTH, HHT, HHH}

\(\therefore\) Total fabourable outcomes = n (E) = 7

\(\therefore\) Probability that at least one head appears is

P = \(\frac {Total \, favourable \, outcome}{Total\, possible \, outcomes}\) = \(\frac 78\)

Correct option is: D) \(\frac{7}{8}\)

Answer is (B) 24

White balls = 4, Black balls = 3

and Red balls = 2

Favourable cases of three different colour balls

= 4 × 3 × 2 = 24

Answer is (B) 2

Total days in a leap year = 366

Remaining day of after 53 weeks

= 366/53 = 2 remaining

P(A ∪ B) = P(A) + P(B)

⇒ P{B) = P(A ∪ B) – P(A)

⇒ P(B) = K = 0.5 – 0.3

= 0.2

Thus K = 0.2.

Answer is (D) Complementary

Answer is (A) P(A) + P(B)

We have, a sample space associated with an experiment is 

S = {(x, y): x = 1,6 and 

y = 1.2,3,4,5,6} 

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

∴ n(S) = 12 

(i) Let A: ‘sum of the numbers that turn up is 3’ 

∴ A = {(1,2)} ∵ 1 + 2 = 3 

⇒ n(A) = 1 

∴ Required probability

= P(A) = n(A)/n(S) = 1/12

 (ii) Let B: ‘sum of the numbers that turn up is 12’. 

⇒ B = {(6,6)} ∵ 6 + 6 = 12 

∴ Required probability

= P(B) = 1/12 = n(B)/n(S)

Number of members on the city council 

= 4 + 6 = 10 

n(S)= ¹⁰C₁ =10 

Let A: ‘selected member is a woman’ 

⇒n(A)= ⁶C₁ = 6 

∴ Required probability

= P(A) = n(A)/n(S) = 6/10 = 3/5

Answer: (c)  \(\frac{3}{7}\) 

P(Drawing a green ball from bag A) = \(P\big(\frac{G}{A}\big) =\frac{4}{7}\)

P(Drawing a green ball from bag B)  =  \(P\big(\frac{G}{B}\big) =\frac{3}{7}\)

∴  Required probability \( P\big(\frac{B}{G}\big)\)

= \(\frac{P(B).P\big(\frac{G}{B}\big)}{P(A).P\big(\frac{G}{A}\big)+P(B).P\big(\frac{G}{B}\big)}\)

\(\frac{\frac{1}{2}.\frac{3}{7}}{\frac{1}{2}.\frac{4}{7}+\frac{1}{2}.\frac{3}{7}}\)  = \(\frac{3/14}{4/14+3/14} = \frac{3/14}{7/14} = \frac{3}{7}\)

Answer is (B) 2ⁿ

Given: A bag containing 6 red, 4 white and 8 blue balls.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Three balls are drawn so, we have to find the probability that one is red, one is white and one is blue.

Total number of outcomes for drawing 3 balls is ¹⁸C₃

n (S) = ¹⁸C₃ = 816

Let E be the event that one red, one white and one blue ball is drawn.

n (E) = ⁶C₁⁴C₁⁸C₁ = 192

P (E) = n (E) / n (S)

= 192 / 816

= 4/17

Given: A bag containing 7 white, 5 black and 4 red balls.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Two balls are drawn at random, therefore

Total possible outcomes are ¹⁶C₂

n (S) = 120

(i) Let E be the event of getting both white balls

E = {(W) (W)}

n (E) = ⁷C₂ = 21

P (E) = n (E) / n (S)

= 21 / 120

= 7/40

(ii) Let E be the event of getting one black and one red ball

E = {(B) (R)}

n (E) = ⁵C₁⁴C₁ = 20

P (E) = n (E) / n (S)

= 20 / 120

= 1/6

(iii) Let E be the event of getting both balls of same colour

E = {(B) (B)} or {(W) (W)} or {(R) (R)}

n (E) = ⁷C₂ + ⁵C₂ + ⁴C₂ = 37

P (E) = n (E) / n (S)

= 37/120

Let A: MBA aspirant will join TIM 

⇒ P(A) = \(\frac{2}{5}\)

B: Will join XLRI ⇒ P(B) = \(\frac{1}{3}\)

A ∩ B = 9 

∴ P(Join hI Mor x LRl) = P(A∪ B) 

= P(A) + (PB) = \(\frac{2}{5}\) + \(\frac{1}{3}\) =  \(\frac{11}{15}\)

P(Neither llM nor x LRI) = P(A’∪ B’) = P(A∪ B)’ = 1 – P(A∪B)

= 1 - \(\frac{11}{15}\) = \(\frac{15 - 11}{15}\)

= \(\frac{4}{15}\)

given: pack of 52 cards from which face cards are removed

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

four cards are drawn from the remaining 40 cards, so we have to find the probability that all of them belong to different suit 

total possible outcomes of drawing four cards are ⁴⁰C₄ 

therefore n(S)= ⁴⁰C₄ 

let E be the event that 4 cards belong to different suit 

n(E)= ¹⁰C₁ ¹⁰C₁ ¹⁰C₁ ¹⁰C₁ = 10000

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{10000}{91390}\) = \(\frac{1000}{9139}\) 

given: box with 100 bulbs of which, 20 are defective

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

ten bulbs are drawn at random for inspection, therefore 

total possible outcomes are ¹⁰⁰C₁₀

therefore n(S)= ¹⁰⁰C₁₀ 

(i) let E be the event that all ten bulbs are defective

n(E)= ²⁰C₁₀

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{20_{c_{10}}}{100_{c_{10}}}\)

(ii) let E be the event that all ten good bulbs are selected

n(E)= ⁸⁰C₁₀

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{80_{c_{10}}}{100_{c_{10}}}\)

(iii) let E be the event that at least one bulb is defective 

E = {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs 

Let E’ be the event that none of the bulb is defective

n(E')= ⁸⁰C₁₀

P(E') = \(\frac{n(E')}{n(S)}\)

P(E') = \(\frac{80_{c_{10}}}{100_{c_{10}}}\)

Therefore, 

P(E) = 1-P(E’)

P(E) = 1-P(E’)

P(E) = \(1-\frac{80_{c_{10}}}{100_{c_{10}}}\)

(iv) let E be the event that none of the selected bulb is defective

n(E)= ⁸⁰C₁₀

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{80_{c_{10}}}{100_{c_{10}}}\)

Given: A bag containing 6 red, 4 white and 8 blue balls.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Two balls are drawn at random.

Total possible outcomes are ¹⁸C₃

n (S) = 816

(i) Let E be the event of getting one red and two white balls

E = {(W) (W) (R)}

n (E) = ⁶C₁⁴C₂ = 36

P (E) = n (E) / n (S)

= 36 / 816

= 3/68

(ii) Let E be the event of getting two blue and one red

E = {(B) (B) (R)}

n (E) = ⁸C₂⁶C₁ = 168

P (E) = n (E) / n (S)

= 168 / 816

= 7/34

(iii) Let E be the event that one of the balls must be red

E = {(R) (B) (B)} or {(R) (W) (W)} or {(R) (B) (W)}

n (E) = ⁶C₁⁴C₁⁸C₁ + ⁶C₁⁴C₂ + ⁶C₁⁸C₂ = 396

P (E) = n (E) / n (S)

= 396 / 816

= 33/68