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New answer posted•a year ago

Probability Class 11th

In a bombing attack, there is 50% chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that there is at least 99% chance of completely destroying the target, is

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

11.00
Let probability of hitting the target = p ⇒ p=1/2
Let n be the minimum number of bombs
According to given condition
1 - (?C?P?(1-P)? + ?C?P¹(1-P)?¹) ≥ 99/100
⇒ 2? ≥ (n+1)100
n=10 ⇒ 2¹? ≥ 1100 Reject
n=11 ⇒ 2¹¹ ≥ 1200 Select

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter One

The mean and variance of a binomial distribution are and $\frac{α}{3}$ respectively. If P(X = 1) = $\frac{4}{243}$ then P(X = 4 or 5) is equal to:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Given, mean = np = . and variance = npq = $\frac{α}{3} \Rightarrow q = \frac{1}{3} a n d p = \frac{2}{3}$

$P (X = 1) = n p^{1} q^{n - 1} = \frac{4}{243} \Rightarrow n {(\frac{2}{3})}^{1} {(\frac{1}{3})}^{n - 1} = \frac{4}{243} \Rightarrow n = 6$

$P (X = 4 o r 5) =^{6} C_{4} {(\frac{2}{3})}^{4} {(\frac{1}{3})}^{2} +^{6} C_{5} {(\frac{2}{3})}^{5} {(\frac{1}{3})}^{1} = \frac{16}{27}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Let X be a binomially distributed random variable with mean 4 and variance $\frac{4}{3} .$ Then, $54 P (X \leq 2)$ is equal to

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Mean = 4 = $μ$ = np

Variance = $σ^{2} = n p (1 - p) = \frac{4}{3} \Rightarrow 4 (1 - p) = \frac{4}{3} \Rightarrow p = \frac{2}{3} \Rightarrow n = 6$

$=^{6} C_{0} {(\frac{1}{3})}^{6} +^{6} C_{1} {(\frac{2}{3})}^{1} {(\frac{1}{3})}^{6} +^{6} C_{2} {(\frac{2}{3})}^{2} {(\frac{1}{3})}^{4} = \frac{146}{27}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If $σ^{2}$ is the variance of X, then $100 σ^{2}$ is equal to………

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

X 0 1 2 3

P (X) $\frac{1}{6}$ $\frac{1}{2}$ $\frac{3}{10}$ $\frac{1}{30}$

$σ^{2} = Σ x^{2} P (x) - (Σ \times P {(x)}^{2} = \frac{56}{100})$

$100 σ^{2} = 56$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Let A and B be two events such that $P (A | B) = \frac{2}{5}, P (A | B) = \frac{1}{7} a n d P (A \cap B) = \frac{1}{9} .$ Consider $(S 1) P (A' \cup B) = \frac{5}{6},$

$(S 2) P (A' \cap B') = \frac{1}{18}$

Then

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$P (A / B) = \frac{1}{7} \Rightarrow \frac{P (A \cap B)}{P (B)} = \frac{1}{7}$

$\Rightarrow P (B) = \frac{7}{9}$

$P (B / A) = \frac{2}{5} \Rightarrow \frac{P (A \cap B)}{P (A)} = \frac{2}{5}$

$\Rightarrow P (A) = \frac{5}{18}$

Now, $P (A' \cup B) = 1 - P (A \cup B) + P (B)$

Both (S1 ) and (S2 ) are true

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The number of positive integers k such that the constant term in the binomial expansion of ${(2 x^{3} + \frac{3}{x^{k}})}^{12}, x \neq 0 i s 2^{8} . l, w h e r e l$  is an odd integer, is……….

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

${(2 x^{3} + \frac{3}{x^{k}})}^{12}$

gen term =

$=^{12} C_{r} 2^{12 - r} . 3^{r} . x^{36 - 3 r - r k}$

For constant term

36 – 3r – rk = 0

$\Rightarrow k = \frac{36 - 3 r}{r}$

for r = 1, 2, 4

¹²C_r2^12-r>2⁸

Possible values of k = 3, 1

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The probability, that in a randomly selected 3-digit number at least two digits are odd, is

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Let $S = {E_{1}, E_{2}, . . . . . ., E_{8}}$  be a sample space of a random experiment such that P(E_n) = $\frac{n}{36}$  for every n = 1, 2,……, 8. then the number of elements in the set  ${A \subseteq S : P (A) \geq \frac{4}{5}}$ is……….

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

P (E_n) = n/36 for n = 1, 2, 3, …., 8

$P (A) = \frac{A n y p o s s i b l e s u m o f (1, 2, 3, . . . . . . . . ., 8) (= α s a y)}{36}$

$\frac{α}{36} \geq \frac{4}{5}$

$∴ a \geq 29$

If one of the number from {1, 2, ….8} is left then total $\geq$ 29 by 3 ways

Similarly by leaving terms more 2 or 3 we get 16 more combinations

$∴$ Total number of different set a possible is 16 + 3

= 19

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

If a point A(x, y) lies in the region bounded by the y-axis, straight lines 2y + x = 6 and 5x – 6y = 30, then the probability that y < 1 is

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

The required probability

$= \frac{A r e a o f R e g i o n P Q C A P}{A r e a o f R e g i o n A B C A}$

$= \frac{\frac{1}{2} \times 8 \times 6 - \frac{1}{2} \times 2 \times 4}{\frac{1}{2} \times 8 \times 6}$

$= \frac{5}{6}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The sum of probabilities of two students getting distinction in their final examinations is 1.2.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} Since, t h e t w o g i v e n e v e n t s a r e n o t r e l a t e d t o t h e s a m e S a m p l e s p a c e . \\ ∴ T h e s u m o f p r o b a b i l i t i e s o f t w o s t u d e n t s g e t t i n g d i s t i n c t i o n i n t h e i r f i n a l \\ m a y b e 1.2 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' \end{array}$

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