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New answer posted•a year ago

Probability Class 12th

9. Determine: Mother, father and son line up at random for a family picture.

E : Son on one end, F : Father in middle.

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

9. The sample space of given condition is

$S = {(M F S, M S F, F M S, F S M, S M F, S F M)}$

$E : {M F S, F M S, S M F, S F M}$

$F : {M F S, S F M}$

$P (F) = \frac{2}{6} = \frac{1}{3}$ , $P (E \cap F) = \frac{2}{6} = \frac{1}{3}$

$P (E / F) = \frac{P (E \cap F)}{P (F)} = \frac{\frac{1}{3}}{\frac{1}{3}} = 1$

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New answer posted•a year ago

Probability Class 12th

8. Determine: E A dice is thrown three times.

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.

13 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

If a die is thrown three times, then the number of elements in the sample space will be $6 \times 6 \times 6 = 216$

Here, $E =$ ${\begin{cases} (1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4) \\ (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4) \\ (3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6, 4) \\ (4, 1, 4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4) \\ (5, 1, 4), (5, 2, 4), (5, 3, 4), (5, 4, 4), (5, 5, 4), (5, 6, 4) \\ (6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4) \end{cases}}$

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

$? E ? F = {(6, 5, 4)}$

$i . e ., P (F) = \frac{6}{216}$ $, P (E ? F) = \frac{1}{216}$

$P (E / F) = \frac{P (E ? F)}{P (F)} = \frac{\frac{1}{216}}{\frac{6}{216}} = \frac{1}{6}$

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New answer posted•a year ago

Probability Class 12th

7. Determine :Two coins are tossed once.

(i) E : tail appears on one coin, F : one coin shows head.

(ii) E : no tail appears, F : no head appears.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

7. The sample space of given experiment is

$S = {H H, H T, T H, T T}$

$(i) E : {H T, T H}$

$\Rightarrow F : {H T, T H}$

$\Rightarrow E \cap F = {H T, T H}$

$P (F) = \frac{2}{4} = \frac{1}{2}$ $; P (E \cap F) = \frac{2}{4} = \frac{1}{2}$

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

$(ii) E : {H H}$

$F : {T T}$

$E \cap F : {}$

So, $, P (F) = \frac{1}{4}$ $, P (E \cap F) = \frac{0}{4} = 0$

$P (E / F) = \frac{P (E \cap F)}{P (F)} = \frac{0}{\frac{1}{4}} = 0$

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New answer posted•a year ago

Probability Class 12th

6. Determine: A coin is tossed three times.

(i) E : heads on third toss, F : heads on first two tosses.

(ii) E : at least two heads, F : at most two heads.

(iii) E : at most two tails, F : at least one tail.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

6. The sample space has 8 outcome,

$S = {H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T}$

$(i) E = {H H H, H T H, T H H, T T H}$

$F = {H H H, H H T}$

and $E \cap F = {H H H}$

$P (F) = \frac{2}{8}$ $; P (E \cap F) = \frac{1}{8}$

$P (E / F) = \frac{P (E \cap F)}{P (F)} = \frac{\frac{1}{8}}{\frac{2}{8}} = \frac{1}{2}$

$(ii) E = {H H H, H H T, H T H, T H H}$

$F = {H H T, H T H, H T T, T H H, T H T, T T H, T T T}$

$E \cap F = {H H H, H T H, T H H}$

$P (F) = \frac{7}{8}$ $; P (E \cap F) = \frac{3}{8}$

$P (E / F) = \frac{P (E \cap F)}{P (F)} = \frac{\frac{3}{8}}{\frac{7}{8}} = \frac{3}{7}$

$(iii) E : {H H H, H H T, H T T, H T H, T H H, T H T, T T H}$

$F : {H H T, H T H, H T T, T H H, T H T, T T T, T T H}$

$E \cap F : {H H T, H T H, H T T, T H H, T H T, T T H}$

$P (F) = \frac{7}{8}$ $; P (E \cap F) = \frac{6}{8}$

$P (E / F) = \frac{P (E \cap F)}{P (F)} = \frac{\frac{5}{8}}{\frac{6}{8}} = \frac{6}{7}$

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New answer posted•a year ago

Probability Class 12th

5. If P(A)=6/11, P(B) =5/11 and P(A ∪ B) =7/11, find

(i) P(A ∩ B)

(ii) P(A|B)

(iii) P(B|A)

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

5. Given,

$P (A) = \frac{6}{11}$

$P (B) = \frac{5}{11}$

$P (A \cup B) = \frac{7}{11}$

$(i) P (A \cap B)$

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$\Rightarrow \frac{7}{11} = \frac{6}{11} + \frac{5}{11} - P (A \cap B)$

$\Rightarrow \frac{7}{11} - \frac{11}{11} = - P (A \cap B)$

$\Rightarrow - \frac{4}{11} = - P (A \cap B)$

$\Rightarrow P (A \cap B) = \frac{4}{11}$

$(ii) P (A / B)$

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

$= \frac{\frac{4}{11}}{\frac{5}{11}} = \frac{4}{5}$

$(iii) P (B / A)$

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

$= \frac{\frac{4}{11}}{\frac{6}{11}} = \frac{4}{6} = \frac{2}{3}$

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New answer posted•a year ago

Probability Class 12th

4. Evaluate P (A ∪ B), if 2P (A) = P (B) =5/13 and P(A|B) =2/5

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

4. Given,

$2 P (A) = P (B) = \frac{5}{13}$

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

$2 P (A) = \frac{5}{13} \Rightarrow P (A) = \frac{5}{13 \times 2} = \frac{5}{26}$

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

$\Rightarrow \frac{2}{5} = \frac{P (A \cap B)}{\frac{5}{13}}$

$\Rightarrow \frac{2}{5} \times \frac{5}{13} = P (A \cap B) \Rightarrow P (A \cap B) = \frac{2}{13}$

$∴ P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$= \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$

$= \frac{5 + 10 - 4}{26} = \frac{11}{26}$

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New answer posted•a year ago

Probability Class 12th

3. If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find

(i) P(A ∩ B)

(ii) P(A|B)

(iii) P(A ∪ B)

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

3. Given,

$P (A) = 0.8$

$P (B) = 0.5$

$P (B / A) = 0.4$

$(i) P (A \cap B)$

$P (B / A) = \frac{P (B \cap A)}{P (A)}$ $[? P (A \cap B) = P (B \cap A)]$

$\Rightarrow 0.4 = \frac{P (A \cap B)}{0.8}$

$\Rightarrow P (A \cap B) = 0.4 \times 0.8$

$= 0.32$

$(ii) P (A / B)$

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

$P (A / B) = \frac{0.32}{0.5}$

$= 0.64$

$(iii) P (A \cup B)$

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$= 0.8 + 0.5 - 0.32$

$= 0.98$

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New answer posted•a year ago

Probability Class 12th

2. Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

2. Given,

$P (B) = 0.5$

$P (A \cap B) = 0.32$

$P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{0.32}{0.5} = \frac{32}{50} = \frac{16}{45}$

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New answer posted•a year ago

Probability Class 12th

1. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E).

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

1. Given,

$P (E) = 0.6$

$P (F) = 0.3$

$P (E \cap F) = 0.2$

$P (E / F) = \frac{P (E \cap F)}{P (F)} = \frac{0.2}{0.3} = \frac{2}{3}$

$P (F / E) = \frac{P (F \cap E)}{P (E)} = \frac{0.2}{0.6} = \frac{2}{6} = \frac{1}{3}$

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New answer posted•a year ago

Probability Class 11th

54. The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a Person getting the right sequence to open the suitcase?

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

54. Since, the lock can be open by a combination of four digits from the given ten digits I e, from 0 to 9. The number of ways of selecting 4 digits, = 10C₄

This combination of 4 digits can again be arranged within themselves in 41 ways. So, total number of

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