2 Marks - Page 2 - Maths STD 12 Science Questions - Vidyadip

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2 Marks

Question 512 Marks

Three numbers are chosen from 1 to 20. Find the probability that they are consecutive.

Answer

S = There numbers are chosen from 1 to 20
n(S) = ²⁰C₃
E = Group of three consecutive numbers between 1 and 20
n(E) = 18
{(1, 2, 3,), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7), (6, 7, 8), (7, 8, 9), (8, 9, 10), (9, 10, 11), (10, 11, 12), (11, 12, 13), (11, 12, 13), (12, 13, 14), (13, 14, 15), (14, 15, 16), (15, 16, 17), (16, 17, 18), (17, 18, 19), (18, 19, 20)}
$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}$
$=\frac{18}{^{20}\text{C}_3}$
Required probability $=\frac{18}{^{20}\text{C}_3}$

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Question 522 Marks

Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima's selection is $\frac{1}{7}$ and that of John's selection is $\frac{1}{5}$. What is the probability that,
None of them will be selected?

Answer

Given,
Probability of Fatima's (F) selection $=\frac{1}{7}$
$\text{P(F)}=\frac{1}{7}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{6}{7}$
Probability of John's (J) selection $=\frac{1}{5}$
$\text{P(F)}=\frac{1}{5}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{4}{5}$
P(None of them selected)
$=\text{P}(\overline{\text{F}}\cap\overline{\text{J}})$
$=\text{P}(\overline{\text{F}})+\text{P}(\overline{\text{J}})$
$=\frac{6}{7}\times\frac{4}{5}$
$=\frac{24}{35}$
Required probabilty $=\frac{24}{35}$

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Question 532 Marks

Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
The first is a king and the second is an ace.

Answer

We know that, there are 4 king and 4 ace in a pack of 52 cards.
Two cards are drawn without replacment
A = First catd is king
B = Second card an ace
P (The first card is a king second is an ace)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{4}{52}\times\frac{4}{21}$
$=\frac{4}{663}$
Required probability $=\frac{4}{663}$

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Question 542 Marks

A husband and wife appear in an interview for two vacancies for the same post. The probability of husband's selection is $\frac{1}{7}$ and that of wife's selection is $\frac{1}{5}$. What is the probability that,
Both of them will be selected?

Answer

Given,
Probability of Husband's (H) selection $=\frac{1}{7}$
$\text{P(H)}=\frac{1}{7}$
Probability of wife's (W) selection $=\frac{1}{5}$
$\text{P(W)}=\frac{1}{5}$
P(Both of them will be selelcted)
$=(\text{H}\cap\text{W})$
$=\text{P(H)}\text{ P(W)}$
$=\frac{1}{7}\times\frac{1}{5}$
$=\frac{1}{35}$
Required probability $=\frac{1}{35}$

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Question 552 Marks

If $\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{11},$ find
$\text{P}(\text{A}\cap\text{B})$

Answer

Given,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11},\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
Since, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})$
$=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}$
$\text{P}(\text{A}\cap\text{B})=\frac{4}{11}$

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Question 562 Marks

Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$

Answer

Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P(A) }\text{P(B)}$
$=\big[1-\text{P(A)}\big]\big[1-\text{P(B)}\big]$
$=(1-0.3)(1-0.6)$
$=0.7\times0.4$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=0.28$

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Question 572 Marks

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
Both the balls are red.

Answer

Given,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (Both the balls are red)
$=\text{P}(\text{R}_1\cap\text{R}_2)$
$=\text{P}(\text{R}_1)\text{ P}(\text{R}_2)$
$=\frac{8}{18}\times\frac{8}{18}$
$=\frac{16}{81}$
Required probability $=\frac{16}{81}$

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Question 582 Marks

A bag contains 4 red and 5 black balls, a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag, find the probability that the,
Balls are of different colours.

Answer

Given,
Bag (1) contains 4 red and 6 black balls.
Bag (2) contains 3 red and 7 black balls
One ball is drawn ar random from each bag.
P (Balls are of different colours)
$=\text{P}\big((\text{R}_1\cap\text{B}_2)\cup(\text{B}_1\cap\text{R}_2)\big)$
$=\text{P}(\text{R}_1\cap\text{B}_2)+\text{P}(\text{B}_1\cap\text{R}_2)$
$=\text{P}(\text{R}_1)\text{P}(\text{B}_2)+\text{P}(\text{B}_1)\text{P}(\text{R}_2)$
$=\frac{4}{9}\times\frac{7}{10}+\frac{5}{9}\times\frac{9}{10}$
$=\frac{28}{90}+\frac{15}{90}$
$=\frac{43}{90}$

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Question 592 Marks

Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima's selection is $\frac{1}{7}$ and that of John's selection is $\frac{1}{5}$. What is the probability that,
Only one of them will be selected?

Answer

Given,
Probability of Fatima's (F) selection $=\frac{1}{7}$
$\text{P(F)}=\frac{1}{7}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{6}{7}$
Probability of John's (J) selection $=\frac{1}{5}$
$\text{P(F)}=\frac{1}{5}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{4}{5}$
P(Only one of them selected)
$=\text{P}\big((\text{F}\cap\overline{\text{J}})\cup(\overline{\text{F}}\cap\text{J})\big)$
$=\text{P}(\text{F}) \text{P}(\overline{\text{J}})+\text{P}(\overline{\text{F}})\text{P}(\text{J})$
$=\frac{1}{7}\times\frac{4}{5}+\frac{6}{7}\times\frac{1}{5}$
$=\frac{4+6}{35}$
$=\frac{10}{35}$
$=\frac{2}{7}$
Required probability $=\frac{2}{7}$

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Question 602 Marks

Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$

Answer

Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{0.18}{0.3}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.6$

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Question 612 Marks

If A and B are two events such that $\text{P}(\text{A}\cap\text{B})=0.32$ and P (B) = 0.5, find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$.

Answer

Given:
$\text{P}(\text{A}\cap\text{B})=0.32$ and $\text{P(B)}=0.5$
We know that,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.32}{0.5}$
$=\frac{16}{25}$
$=0.64$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.64$

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Question 622 Marks

A card is drawn from a well-shulffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.
What is the probability that the first card is an ace and the second card is a red queen?

Answer

P(first ace and second red queen) = P(ace card) × P(red queen)
$=\frac{4}{52}\times\frac{2}{52}$
$=\frac{1}{338}$

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Question 632 Marks

A bag contains 4 red and 5 black balls, a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag, find the probability that the,
Balls are of the same colour.

Answer

Given,
Bag (1) contains 4 red and 6 black balls.
Bag (2) contains 3 red and 7 black balls
One ball is drawn ar random from each bag.
P (Balls are of the same colour)
$=\text{P}\big((\text{B}_1\cap\text{B}_2)\cup(\text{R}_1\cap\text{R}_2)\big)$
$=\text{P}(\text{B}_1\cap\text{B}_2)+\text{P}(\text{R}_1\cap\text{R}_2)$
$=\text{P}(\text{B}_1)\text{P}(\text{B}_2)+\text{P}(\text{R}_1)\text{P}(\text{R}_2)$
$=\frac{5}{9}\times\frac{7}{10}+\frac{4}{9}\times\frac{3}{10}$
$=\frac{35}{90}+\frac{12}{90}$
$=\frac{47}{90}$
Required probability $=\frac{47}{90}$

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Question 642 Marks

Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\overline{\text{A}}\cap\text{B})$

Answer

Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.6 - 0.18$
$\text{P}(\overline{\text{A}}\cap\text{B})=0.42$

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Question 652 Marks

Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal's selection is $\frac{1}{3}$ and that of Monika's selection is $\frac{1}{5}$. Find the probability that.
none of them will be selected.

Answer

P(Kamal gets selected) $=\text{P(A)}=\frac{1}{3}$
P(Monica gets selected) $=\text{P(B)}=\frac{1}{5}$
P(none of them get selected) $=\text{P}(\overline{\text{A}})\times\text{P}(\overline{\text{B}})$
$=[1-\text{P(A)}][1-\text{P(B)}]$
$=\Big(1-\frac{1}{3}\Big)\Big(1-\frac{1}{5}\Big)$
$=\frac{2}{3}\times\frac{4}{5}$
$=\frac{8}{15}$

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2 Marks - Page 2 - Maths STD 12 Science Questions - Vidyadip