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Class 12th Probability

A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted

a year ago

Class 12th Probability

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, n = 7, p = 0.25 = \frac{25}{100} = \frac{1}{4}, q = 1 - \frac{1}{4} = \frac{3}{4} \\ ∴ P (X \geq 2) = 1 - [P (x = 0) + P (x = 1)] \\ = 1 - [{}_{7}{C_{0}} {(\frac{1}{4})}^{0} {(\frac{3}{4})}^{7} + {}_{7}{C_{1}} {(\frac{1}{4})}^{1} {(\frac{3}{4})}^{6}] \\ = 1 - [{(\frac{3}{4})}^{7} + \frac{7}{4} {(\frac{3}{4})}^{6}] = 1 - {(\frac{3}{4})}^{6} [\frac{3}{4} + \frac{7}{4}] \\ = 1 - {(\frac{3}{4})}^{6} (\frac{10}{4}) = 1 - \frac{729}{4096} * \frac{10}{4} = 1 - \frac{7290}{16384} \\ = \frac{16384 - 7290}{16384} = \frac{9094}{16384} = \frac{4547}{8192} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{4547}{8192} . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Ten coins are tossed. What is the probability of getting at least 8 heads?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted

a year ago

Class 12th Probability

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, p = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2} \\ \Rightarrow q = 1 - \frac{1}{2} = \frac{1}{2} a n d n = 5 \\ ∴ P (x = r) = {}_{n}{C_{r}} p^{r} q^{n - r} \\ = {}_{5}{C_{3}} {(\frac{1}{2})}^{3} {(\frac{1}{2})}^{5 - 3} = \frac{5!}{3! 2!} . {(\frac{1}{2})}^{3} . {(\frac{1}{2})}^{2} = 10 . \frac{1}{8} . \frac{1}{4} = \frac{5}{16} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{5}{16} . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are kings?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1}, E_{2}, E_{3} a n d E_{4} b e t h e e v e n t s t h a t I, I I, I I I a n d I V c a r d i s K i n g r e s p e c t i v e l y . \\ ∴ P (E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) \\ = P (E_{1}) . P (E_{2} / E_{1}) . P [\frac{E_{3}}{E_{1} \cap E_{2}}] . P [\frac{E_{4}}{(E_{1} \cap E_{2} \cap E_{3} \cap E_{4})}] \\ = \frac{4}{52} * \frac{3}{51} + \frac{2}{50} * \frac{1}{49} = \frac{24}{52.51.50.49} = \frac{1}{13.17.25.49} = \frac{1}{27075} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{1}{27075} . \end{array}$

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New answer posted

a year ago

Class 12th Probability

A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of the second ball being blue?

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t t h e b o x h a s 5 b l u e a n d 4 r e d b a l l s . \\ L e t E_{1} = T h e e v e n t t h a t I b a l l d r a w n i s b l u e \\ E_{2} = T h e e v e n t t h a t I b a l l d r a w n i s r e d \\ a n d E = T h e e v e n t t h a t a I I b a l l d r a w n i s b l u e \\ ∴ P (E_{1}) = \frac{5}{9}, P (E_{2}) = \frac{4}{9}, P (E / E_{1}) = \frac{4}{8} a n d P (E / E_{2}) = \frac{5}{8} \\ ∴ P (E) = P (E_{1}) . P (E / E_{1}) + P (E_{2}) . P (E / E_{2}) \\ = \frac{5}{9} * \frac{4}{8} + \frac{4}{9} * \frac{5}{8} = \frac{20}{72} + \frac{20}{72} = \frac{40}{72} = \frac{5}{9} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{5}{9} . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Bag I contains 3 black and 2 white balls; Bag II contains 2 black and 4 white balls. A bag and a ball are selected at random. Determine the probability of selecting a black ball.

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t b a g I = {3 B, 2 W} \\ a n d b a g I I = {2 B, 4 W} \\ L e t E_{1} = T h e e v e n t t h a t b a g I i s s e l e c t e d \\ E_{2} = T h e e v e n t t h a t b a g I I i s s e l e c t e d \\ a n d E = T h e e v e n t t h a t a b l a c k b a l l i s s e l e c t e d \\ ∴ P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{2}, P (E / E_{1}) = \frac{3}{5} a n d P (E / E_{2}) = \frac{1}{3} \\ ∴ P (E) = P (E_{1}) . P (E / E_{1}) + P (E_{2}) . P (E / E_{2}) \\ = \frac{1}{2} * \frac{3}{5} + \frac{1}{2} * \frac{1}{3} = \frac{3}{10} + \frac{1}{6} = \frac{9 + 5}{30} = \frac{14}{30} = \frac{7}{15} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{7}{15} . \end{array}$

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New answer posted

a year ago

Class 12th Probability

A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t W_{1} a n d W_{2} b e t w o b a g s c o n t a i n i n g (4 W, 5 B) a n d (9 W, 7 B) b a l l s r e s p e c t i v e l y . \\ L e t E_{1} b e t h e e v e n t t h a t t h e t r a n s f e r r e d b a l l f r o m t h e b a g W_{1} t o W_{2} i s w h i t e a n d \\ E_{2} b e t h e e v e n t t h a t t h e t r a n s f e r r e d b a l l i s b l a c k . \\ a n d E b e t h e e v e n t t h a t t h e b a l l d r a w n f r o m t h e second b a g i s w h i t e . \\ ∴ P (E_{1}) = \frac{4}{9}, P (E_{2}) = \frac{5}{9}, P (E / E_{1}) = \frac{10}{17} a n d P (E / E_{2}) = \frac{9}{17} \\ ∴ P (E) = P (E_{1}) . P (E / E_{1}) + P (E_{2}) . P (E / E_{2}) \\ = \frac{4}{9} * \frac{10}{17} + \frac{5}{9} * \frac{9}{17} = \frac{40}{153} + \frac{45}{153} = \frac{85}{153} = \frac{5}{9} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{5}{9} . \end{array}$

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New answer posted

a year ago

Class 12th Probability

Suppose 10,000 tickets are sold in a lottery, each for Re 1. First prize is of Rs 3000 and the second prize is of Rs 2000. There are three third prizes of Rs 500 each. If you buy one ticket, what is your expectation?

0 Follower 12 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

0 0

New answer posted

a year ago

Class 12th Probability

Three dice are thrown at the same time. Find the probability of getting three twos, if it is known that the sum of the numbers on the dice was six.

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e d i c e i s t h r o w n t h r e e t i m e s \\ ∴ S a m p l e s p a c e n (S) = {(6)}^{3} = 216 \\ L e t E_{1} b e t h e e v e n t w h e n t h e s u m o f n u m b e r s o n t h e d i c e w a s s i x a n d E_{2} b e t h e e v e n t w h e n \\ t h r e e t w o' s o c c u r . \\ \Rightarrow E_{1} = {(1, 1, 4), (1, 2, 3), (1, 3, 2), (1, 4, 1), (2, 1, 3), (2, 2, 2), (2, 3, 1), (3, 1, 2), (3, 2, 1), (4, 1, 1)} \\ \Rightarrow n (E_{1}) = 10 a n d n (E_{2}) = 1 [? E_{2} = {2, 2, 2}] \\ ∴ P (E_{2} / E_{1}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{1 / 216}{10 / 216} = \frac{1}{10} . \end{array}$

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