Tags Probability

Probability

Get insights from 306 questions on Probability, answered by students, alumni, and experts. You may also ask and answer any question you like about Probability

Follow Ask Question

306

Questions

Discussions

Active Users

Followers

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

The probability distribution of a discrete random variable $X$ is given below:

$X$ : 2 3 4 5
$P (X)$ : $5 / k, 7 / k, 9 / k, 11 / k$
The value of $k$ is:

(a) 8

(b) 16

(c) 32

(d) 48

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t \sum_{i = 1}^{n} P (X_{i}) = 1 \\ ∴ \frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1 \\ \frac{32}{k} = 1 \Rightarrow k = 32 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons, 4 are swimmers is:

^{(a) 5}C₄ (0.7)⁴ (0.3)

^{(b) 5}C₁ (0.7) (0.3)⁴

^{(c) 5}C₁ (0.7) (0.3)⁴

(d) (0.7)⁴ (0.3)

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

0 0

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, n = 10, p = \frac{1}{2}, q = 1 - \frac{1}{2} = \frac{1}{2} (f o r t r u e / f a l s e q u e s t i o n s) \\ a n d r \geq 8 i . e ., 8, 9, 10 \\ ∴ P (X \geq 8) = P (x = 8) + P (x = 9) + P (x = 10) \\ = {}_{10}{C_{8}} {(\frac{1}{2})}^{8} {(\frac{1}{2})}^{2} + {}_{10}{C_{9}} {(\frac{1}{2})}^{9} (\frac{1}{2}) + {}_{10}{C_{10}} {(\frac{1}{2})}^{10} {(\frac{1}{2})}^{0} \\ = 45 . {(\frac{1}{2})}^{10} + 10 . {(\frac{1}{2})}^{10} + {(\frac{1}{2})}^{10} = {(\frac{1}{2})}^{10} (45 + 10 + 1) \\ = 56 * \frac{1}{1024} = \frac{7}{128} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

0 0

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two cards are drawn from a well-shuffled deck of 52 playing cards with replacement. The probability that both cards are queens is:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} P r o b a b i l i t y o f g e t t i n g Q u e e n = \frac{4}{52} \\ S o, t h e r e q u i r e d p r o b a b i l i t y = P (Q u e e n) . P (Q u e e n) \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{4}{52} . \frac{4}{52} = \frac{1}{13} . \frac{1}{13} (w i t h r e p l a c e m e n t) \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

0 0

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Which one is not a requirement of a binomial distribution?

(a) There are 2 outcomes for each trial

(b) There is a fixed number of trials

(c) The outcomes must be dependent on each other

(d) The probability of success must be the same for all the trials

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t f o r a B i n o m i a l d i s t r i b u t i o n, t h e o u t c o m e s m u s t n o t b e d e p e n d e n t o n e a c h o t h e r . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two dice are thrown. If it is known that the sum of the numbers on the dice was less than 6, the probability of getting a sum of 3, is

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t s h o w i n g t h e s u m o f t h e n u m b e r s o n t h e t w o d i c e w a s l e s s t h a n 6 \\ ∴ E_{1} = {(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (2, 2), (2, 3), (3, 2)} \\ n (E_{1}) = 10 \\ . E_{2} b e t h e e v e n t t h a t t h e s u m o f t h e n u m b e r s i s 3 . \\ ∴ E_{2} = {(1, 2), (2, 1)} \\ n (E_{2}) = 2 a n d n (E_{1} \cap E_{2}) = 2 \\ ∴ P (E_{2} / E_{1}) = \frac{n (E_{1} \cap E_{2})}{n (E_{1})} = \frac{2}{10} = \frac{1}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Eight coins are tossed together. The probability of getting exactly 3 heads is

0 Follower 1 View

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

0 0

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} p r o b a b i l i t y = P (d e a d) . P (d e a d) \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{3}{8} . \frac{2}{7} = \frac{3}{28} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A box contains 3 orange balls, 3 green balls, and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} P robability o f d r a w i n g 2 g r e e n a n d 1 b l u e b a l l s \\ = P (G) . P (G) . P (B) + P (G) . P (B) . P (G) + P (B) . P (G) . P (G) \\ = \frac{3}{8} . \frac{2}{7} . \frac{2}{6} + \frac{3}{8} . \frac{2}{7} . \frac{2}{6} + \frac{2}{8} . \frac{3}{7} . \frac{2}{6} = \frac{12}{336} + \frac{12}{336} + \frac{12}{336} = \frac{36}{336} = \frac{3}{28} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

0 0

New answer posted

3 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t o f g e t t i n g e v e n n u m b e r o n t h e d i e . E_{2} b e t h e e v e n t o f s e l e c t i n g a s p a d e c a r d . \\ ∴ P (E_{1}) = \frac{3}{6} = \frac{1}{2} a n d P (E_{2}) = \frac{13}{52} = \frac{1}{4} \\ S o, P (E_{1} \cap E_{2}) = P (E_{1}) . P (E_{2}) = \frac{1}{2} . \frac{1}{4} = \frac{1}{8} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Probability

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience