2 Marks Questions - Page 2 - Maths STD 11 Science Questions - Vidyadip

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2 Marks Questions

Question 512 Marks

A pair of dice is rolled. If the outcome is a doublet, a coin is tossed. Determine the total number of elementary events associated to this experiment.

Answer

If a pair of dices is thrown simultaneously, then all possible outcomes = 6 × 6 = 36.
The set of these outcomes is the sample space, which is given by,

S = { (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Again, if the outcome is a doublet, then a coin is tossed.
Now, we have the following events:
{(1, 1, H), (2, 2, H), (3, 3, H), (4, 4, H), (5, 5, H), (6, 6, H), (1, 1, T), (2, 2, T), (3, 3, T), (4, 4, T), (5, 5, T), (6, 6, T)}
Total number of events when the outcome is a doublet = 6 × 2 = 12
Hence, the total number of elementary events associated with this experiment = (36 - 6) + 12 = 42.

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Question 522 Marks

There are three coloured dice of red, white and black colour. These dice are placed in a bag. One die is drawn at random from the bag and rolled its colour and the number on its uppermost face is noted. Describe the sample space for this experiment.

Answer

A dice has six faces that are numbered from 1 to 6, with one number on each face.
Let us denote the red, white and black dices as R, W, and B, respectively.
Accordingly, when a dice is selected and then rolled, the sample space is given by,

S = {(R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6), (B, 1), (B, 2), (B, 3), (B, 4), (B, 5), (B, 6), (W, 1), (W, 2), (W, 3), (W, 4), (W, 5), (W, 6)}.

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Question 532 Marks

In a single throw of a die describe the following events:
E = Getting an even number greater than 4

Answer

When a dice is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}.
Accordingly, we have:
E = {6}
Here, D = {1, 2, 3} and F = {3, 4, 5, 6}
$\therefore$ D ∩ F = {3}

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Question 542 Marks

Given two mutually exclusive events A and B such that $\text{P(A)} = \frac{1}{2}$ and $\text{P(B)} = \frac{1}{3},$ find P(A or B).

Answer

Given,
$\text{P}(\text{A})=\frac{1}{2}$
$\text{P}(\text{B})=\frac{1}{3}$
$\therefore$ A and B are mutually exclusive events, then $\text{P}(\text{A}\cap\text{B})=0$
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})$
$=\frac{1}{2}+\frac{1}{3}$
$=\frac{3+2}{6}$
$=\frac{5}{6}$
$\therefore\text{P}(\text{A}\cup\text{B})=\frac{5}{6}$

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Question 552 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:Not a black card

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$A black card
Let E be the events that A black card does not appears, which are 26 in numbers (Heart & Diamond)
$\because$ There are 26 black cards (spade and club)
$\therefore\text{n}\text{(E)}=\ ^{26}\text{C}_1=26$ $\text{P}\text{(E)}=\frac{26}{52}=\frac{1}{2}$

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Question 562 Marks

A card is picked up from a deck of 52 playing cards.
What is the sample space of the experiment?

Answer

Sample space for a card picked up from a deck of 52 playing card,
S = {A♠, 2♠, 3♠, 4♠, 5♠, 6♠, 7♠, 8♠, 9♠, 10♠, J♠, Q♠, K♠, A♡, 2♡, 3♡, 4♡, 5♡, 6♡, 7♡, 8♡, 9♡, 10♡, J♡, Q♡, K♡, A♣, 2♣, 3♣, 4♣, 5♣, 6♣, 7♣,
8♣, 9♣, 10♣, J♣, Q♣, K♣, A♢, 2♢, 3♢, 4♢, 5♢, 6♢, 7♢, 8♢, 9♢, 10♢, J♢, Q♢, K♢}

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Question 572 Marks

What is the probability that a leap year has 53 sundays 53 mondays?

Answer

Since ina leap year, there are 52 weeks and two days,
The sample space for the two days will be,
$\text{S} =\big\{\text{(M, T), (T, W), (W, TH), (TH, F), (F, S), (S, SU), (SU, M)}\big\}$
$\therefore\text{n(S)}=7$
$\text{E }= \big\{\text{SU, M}\big\}$
$\Rightarrow\text{n(E)}=1$
$\text{p(E)}=\frac{1}{7}$

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Question 582 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:Neither a heart nor a king

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$Neither a heart nor a king
Let E be the events Neither an ace nor a king appears,
$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the events that either a heart or a king appears
$\therefore\text{n}(\stackrel{{\sim}}{\hbox{E}})=\ ^{6}\text{C}_1+\ ^{4}\text{C}_1+\ ^{1}\text{C}_1 $ =13 + 4 - 1 = 16 [$\therefore$ There is a heart king so, is deducted] $\therefore\text{P}(\stackrel{{\sim}}{\hbox{E}})=\frac{16}{52}=\frac{4}{13}$ $\therefore\ \text{p}\text{(E)}=1-\text{P}(\stackrel{{\sim}}{\hbox{E}})$ $=1-\frac{4}{13}=\frac{9}{13}$

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Question 592 Marks

In a simultanepous throw of a pair of dice, find the probability of getting:
A sum more than 7

Answer

Let E be the events that an even number on the first dice appear which means any number can be apper on second dice,
$\therefore\text{E} = \big\{(2, 6),\ (3, 5),\ (3, 6),\ (4, 4),\ (4, 5),\ (4, 6),\ (5, 3),\ (5, 4),\ \\\ \text{n(E)=15}\ (5,5),\ (5, 6),\ (6,2),\ (6,3),\ (6,4),\ (6,5),\ (6,6) \big\}$
$\text{p(E)}=\frac{15}{36}=\frac{5}{12}$

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Question 602 Marks

If A and B are two events associated with a random experiment such that P(A) = 0.3, P(B) = 0.4 and $\text{P}(\text{A}\cup\text{B})=0.5,$ find $\text{P}(\text{A}\cap\text{B}).$

Answer

We know by addition theorem on probability
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow0.5=0.3+0.4-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\text{A}\cap\text{B})=0.3+0.4-0.5$
$0.7-0.5$
$=0.2$
$\therefore\text{P}(\text{A}\cap\text{B})=0.2$

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Question 612 Marks

A die is thrown twice Each time the number appearing on it is recorded Describe the following events:
B = Both numbers are even.

Answer

When a dice is thrown twice, we have the following possible outcomes:
B = both numbers are even
= {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
Now, we have:
(A ∩ B) = Φ

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Question 622 Marks

If a letter is chosan at random from the english alphabet, find the probability that the letter is:

A vowel
A consonant

Answer

As the letter is choosen from english alphabet$\therefore\ \text{n}\text{(S)}=26$ $\big[\therefore$ there are $26$ letters in english alphabet$\big]$

Let $E$ be the event that a vowel has been choosen

$\therefore\text{n(E)}=\ ^{5}\text{C}_1$
$\therefore\text{P(E)}=\frac{5}{26}$

Probability that a consonant is choosen

$\Rightarrow\text{P}\bar{\text{(E)}}=1-\text{P(E)}$
$=1-\frac{5}{26}=\frac{21}{26}$

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Question 632 Marks

A bag contains 2 red and white balls. three balls are drawn at random. find the probability that:
All the three balls are white.

Answer

Three balls are draw at random,
$\therefore\ \text{n(S)}={^{13}\text{C}_3}$
(All the three balls are white)
$=\frac{{^5\text{C}_{13}}}{{^{13}{\text{C}}_3}}=\frac{\frac{5!}{3!2!}}{\frac{13!}{3!10!}}=\frac{10}{\frac{13\times12\times11}{6}}=\frac{5}{143}$

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Question 642 Marks

If A and B are two events associated with a random experiment such that P(A) = 0.5, P(B) = 0.3 and $\text{P}(\text{A}\cap\text{B})=0.2,$ find $\text{P}(\text{A}\cup\text{B}).$

Answer

We know by addition theorem on probability
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=0.5+0.3-0.2$
$=0.8-0.2$
$=0.6$
$\therefore\text{P}(\text{A}\cup\text{B})=0.6$

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Question 652 Marks

In a random sampling three items are selected from a lot. Each item is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment.

Answer

Three items are to be selected at random from a lot.
Each item in the lot is tested and classified as defective (D) or non-defective (N).
The sample space of this experiment is given by,
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}.

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Question 662 Marks

An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.

Answer

A dice has six faces that are numbered from 1 to 6, with one number on each face.Here, 2, 4 and 6 are even numbers, while 1, 3 and 5 are odd numbers.
A coin has two faces: a head (H) and a tail (T). Hence, the sample space of this experiment is given by, S = {(2, H), (2, T), (4, H), (4, T), (6, H), (6, T), (1, HH), (1, HT), (1,TH), (1, TT), (3, HH), (3, HT), (3, TH), (3, TT), (5, HH), (5, HT), (5, TH), (5, TT)}.

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Question 672 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:Neither an ace nor a king

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$Neither an ace nor a king
Let E be the events that Neither an ace nor a king appears,
$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the events that Neither an ace nor a king appears,
$\therefore\ \text{n}( \stackrel{{\sim}}{\hbox{E}}) =\ ^{4}\text{C}_1+\ ^{4}\text{C}_1$ $=4+4=8$ $\therefore\ \text{P} (\stackrel{{\sim}}{\hbox{E}})=\frac{8}{52}=\frac{2}{13}$ $\therefore\ \text{P}\text{(E)}=1-\text{P} (\stackrel{{\sim}}{\hbox{E}})$ $=1-\frac{2}{13}=\frac{11}{13}$

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Question 682 Marks

Three coins are tossed together,find the probability of getting:At least one head and one tail.

Answer

Three coins are tossed $\therefore\text{n(s)}=2^3=8$At least one head and one tail
$\therefore\ \text{E}= \big\{\text{HTT, THT, TTH, HHT, HTH, THH}\big\}$ $\text{n(E)}=6$ $\therefore\text{P(E)}=\frac{6}{8}=\frac{3}{4}$ $\text{P(E)}=\frac{3}{4}$

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Question 692 Marks

A bag contains 5 red, 6 white and 9 blue balls. two balls are dreawn at random. what is the probability that both balls are red or both are black?

Answer

Bag: 5 - Red, 6 - White, 7 - Blue,
since two balls are drawn at random,
$\therefore\text{n}\text{(s)}=\ ^{18}\text{C}_2$
Let E be the event that all the both balls are either red or black
$\therefore\text{n}\text{(E)}=\ ^{5}\text{C}_2+\ ^{7}\text{C}_2$
$\therefore\text{P}\text{(s)}=\frac{\ ^{5}\text{C}_2+\ ^{7}\text{C}_2}{\ ^{18}\text{C}_2}$
$=\frac{62}{306}=\frac{31}{153}$

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Question 702 Marks

If A and B are two events associated with a random experiment such that $\text{P}(\text{A}\cup\text{B})=0.8,\ \text{P}(\text{A}\cap\text{B})=0.3$ and $\text{P}(\overline{\text{A}})=0.5,$ find P(B).

Answer

We know,
$\text{P}(\text{A}\cup\text{B})=0.8$
$\text{P}(\text{A}\cap\text{B})=0.3$
$\text{P}(\overline{\text{A}})=0.5$
$\Rightarrow1-\text{P}({\text{A}})=0.5$
$\Rightarrow\text{P}({\text{A}})=1-0.5=0.5$
Now, by adding theorem on probability
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$0.8=0.5+\text{P}(\text{B})-0.3$
$0.8=\text{P}(\text{B})+0.2$
$\text{P}(\text{B})=0.8-0.2$
$=0.6$
$\therefore\text{P}(\text{B})=0.6$

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Question 712 Marks

If $A$ and $B$ be mutually exclusive events associated with a random experiment such that $P(A) = 0.4$ and $P(B) = 0.5,$ then find:

$\text{P}(\text{A}\cup\text{B})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$\text{P}(\overline{\text{A}}\cap{\text{B}})$
$\text{P}({\text{A}}\cap\overline{\text{B}})$

Answer

Given,
$P(A) = 0.4$
$P(B) = 0.5$
$\therefore A$ and $B$ are mutually exclusive events, then $\text{P}(\text{A}\cap\text{B})=0$
Now,

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})$

$=0.4+0.5$
$=0.9$
$\therefore\text{P}(\text{A}\cup\text{B})=0.9$

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$

$=1-0.9$
$=0.1$
$\therefore\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=0.1$

$\text{P}(\overline{\text{A}}\cap{\text{B}})=\text{p}(\text{B})-\text{P}({\text{A}}\cap{\text{B}})$

$=0.5-0$
$\therefore\text{P}(\overline{\text{A}}\cap{\text{B}})=0.5$

$\text{P}({\text{A}}\cap\overline{\text{B}})=\text{P}(\text{A})-\text{P}({\text{A}}\cap{\text{B}})$

$=0.4-0$
$=0.4$
$\therefore\text{P}({\text{A}}\cap\overline{\text{B}})=0.4$

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Question 722 Marks

In a single throw of a die describe the following events:
D = Getting a number less than 4

Answer

When a dice is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}.
Accordingly, we have:
D = {1, 2, 3}
Here, E = {6} and F = {3, 4, 5, 6}
$\therefore$ E ∩ F = {6}

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Question 732 Marks

Three coins are tossed together,find the probability of getting:
Exactiy two heads.

Answer

three coins are tossed
$\therefore\text{n(s)}=2^3=8$
E be the event of getting exactly two heads
$\therefore\ \text{E}= \big\{\text{HHT, HTH, THH}\big\}$
$\therefore\text{n(E)}=3$
$\therefore\text{P(E)}=\frac{3}{8}$

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Question 742 Marks

An experiment consists of tossing a coin and then tossing it second time if head occurs. If a tail occurs on the first toss, then a die is tossed once Find the sample space.

Answer

A coin has two faces: a head (H) and a tail (T).
A dice has six faces that are numbered from 1 to 6, with one number on each face.
Thus, the sample space of the given experiment is given by S = {(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

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Question 752 Marks

Two dice are thrown. find the odds in favour of getting the sum:What are the odds against getting the sum 6?

Answer

Two dice are thrown
$\therefore\text{n(E)}=6^2=36$
E be the event of getting sum as 6 is,
$\text{E}=\big\{(1,5),\ (2,4).\ (3,3),\ (4,2),\ (5,1)\big\}$
$\Rightarrow\text{n(E)}=5$
$\therefore\text{P(E)}=\frac{5}{36}$
Also, $\text{P}\bar{\text{(E)}}=1-\text{P(E)}$
$=\frac{31}{36}$
$\therefore$ Odds in favour of getting sum as 6 is,
$\text{P(E)}:\text{P}\bar{\text{(E)}}=31:5$

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Question 762 Marks

Two dice are thrown. find the odds in favour of getting the sum:
4

Answer

Two dice are thrown
$\therefore\text{n(E)}=6^2=36$
E be the event thet total sum is 4 on two dice
$\text{E}=\big\{(1,3),\ (2,2).\ (3,1)\big\}$
$\Rightarrow\text{n(E)}=3$
$\therefore\text{P(E)}=\frac{3}{36}=\frac{1}{12}$
Also, $\text{P}\bar{\text{(E)}}=1-\text{P(E)}$
$=1-\frac{1}{12}=\frac{11}{12}$
Odds in favour of getting sum as 4 is,
$\text{P(E)}:\text{P}\bar{\text{(E)}}=1:11$

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Question 772 Marks

In a simultanepous throw of a pair of dice, find the probability of getting,Neither a doublet nor a toatl of 10.

Answer

Let E be the events that neither a doublet nor a sum of 10 appear on the faces of the dice.$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the events that neither a doublet nor a sum of 10 appear on the faces of the dice.
$\therefore \ \stackrel{{\sim}}{\hbox{E}}\ =\big\{(1, 1),\ (2, 2),\ (3, 3),\ (4, 6),\ (5, 5),\ (6, 4),\ (6, 6)\big\}$ $\therefore\ \text{n}\stackrel{{\sim}}{\hbox{(E)}}=8$ $ \text{p}\stackrel{{\sim}}{\hbox{(E)}}=\frac{8}{36}=\frac{2}{9}$ $\therefore\ \text{p(E)}=1-\text{p}\stackrel{{\sim}}{\hbox{(E)}}$ $=1-\frac{2}{9}=\frac{7}{9}$

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Question 782 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:A jack, queen or a king.

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$A jack, queen or a king
Let E be the events that a jack, queen or a king appear
$\therefore\text{n}(\text{E})=\ ^{2}\text{C}_1+\ ^{4}\text{C}_1+\ ^{4}\text{C}_1 $ = 4 + 4 + 4 = 12 $\therefore\text{P}(\text{E})=\frac{12}{52}=\frac{3}{13}$

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Question 792 Marks

Three coins are tossed together,find the probability of getting:At least two heads.

Answer

Three coins are tossed
$\therefore\text{n(s)}=2^3=8$
E at least two heads(two or 3 heads)
$\therefore\ \text{E}=\big\{\text{HHH, HHT, THH, HTH}\big\}$
$\text{n(E)}=4$
$\therefore\text{P(E)}=\frac{4}{8}=\frac{1}{2}$
$\text{P(E)}=\frac{1}{2}$

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Question 802 Marks

A card is drawn at random from a pack of 52 cards, find the probability that the card drawn is:Spade or an ace.

Answer

since a card is drawn form a pack of 52 cards. Numbers of elementary events in the sample space $\text{n}(\text{E})=\ ^{52}\text{C}_1=52$Spade or an ace.
Let E be the events that either a Spade or an ace appears,
$\text{n}\text{(E)}=\ ^{13}\text{C}_1+\ ^{4}\text{C}_1-\ ^{1}\text{C}_1$ $=13+4-1=16$ $\therefore\text{P}\text{(E)}=\frac{16}{52}=\frac{4}{13}$

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Question 812 Marks

Two dice are thrown. find the odds in favour of getting the sum:5.

Answer

Two dice are thrown
$\therefore\text{n(E)}=6^2=36$
E be the event of getting sum as 5 is,
$\text{E}=\big\{(1,4),\ (2,3).\ (3,2),\ (4,2)\big\}$
$\Rightarrow\text{n(E)}=4$
$\therefore\text{P(E)}=\frac{4}{36}=\frac{1}{9}$
Also, $\text{P}\bar{\text{(E)}}=1-\text{P(E)}$
$=\frac{8}{9}$
Odds in favour of getting sum as 5 is,
$\text{P(E)}:\text{P}\bar{\text{(E)}}=1:8$

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Question 822 Marks

In a simultanepous throw of a pair of dice, find the probability of getting,Odd number on the first and 6 on the second.

Answer

Let E be the events that an odd number on the first and 6 on the second appear on rthe faces of dice.
$\therefore\ \text{E}=\big\{(1, 6),\ (3, 6),\ (5, 6)\big\}$
$\text{n(E)}=3$
$\therefore\ \text{P(E)}=\frac{3}{36}=\frac{1}{12}$

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Question 832 Marks

Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one byb one without, replacement. find the probability that both the ballsw are of different colours.

Answer

2, White, 3, red, 5, green, 4, black
$\because$ Two balls are drawn
$\therefore\text{n(S)}=\ ^{14}\text{C}_2$
Let E be the event that all balls are of the same colour
$\text{E}=\big\{\text{WW, RR,GG,BB}\big\}$
$\therefore\text{n}\text{(E)}=\ ^{2}\text{C}_2+\ ^{3}\text{C}_2+\ ^{5}\text{C}_2+\ ^{4}\text{C}_2$
$\text{P}\text{(E)}=\frac{\ ^{2}\text{C}_2+\ ^{3}\text{C}_2+\ ^{5}\text{C}_2+\ ^{4}\text{C}_2}{\ ^{14}\text{C}_2}$
$=\frac{40}{182}$ $=\frac{20}{19}$
$\therefore$ Probability that both are of different colour
$\text{p(E)}=1-\text{P(E)}$
$=1-\frac{20}{91}$ $=\frac{71}{91}$ $=0.78$

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Question 842 Marks

Two dice are thrown. The events $A, B, C, D, E$ and $F$ are described as follows:
A = Getting an even number on the first die.
B = Getting an odd number on the first die.
C = Getting at most $5$ as sum of the numbers on the two dice.
D = Getting the sum of the numbers on the dice greater than 5 but less than 10.
E = Getting at least $10$ as the sum of the numbers on the dice.
F = Getting an odd number on one of the dice.
Describe the following events:
$A$ and $B, B$ or $C, B$ and $C, A$ and $E, A$ or $F, A$ and $F$.

Answer

When two dices are thrown, there are $6^2 = 36$ possible outcomes.
$A$ = Getting an even number on the first dice
$= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),$
$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }$
$B$ = Getting an odd number on the first dice
$= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),$
$(3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) }$
$C$ = Getting at most 5 as the sum of the numbers on the two dices.
$= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}$
$D$ = Getting a sum greater than 5 but less than 10
$= {(1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6),$
$(4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3)}$
$E$ = Getting at least 10 as the sum of the numbers on the dices
$= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}$
$F$ = Getting an odd number on one of the dices
$= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6),$
$(4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}$
Now,
$A$ and $B = A ∩ B = Φ$
$B$ or $C = B ∪ C$
$= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3),$
$(3, 4), (3, 5), (3, 6), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) }$
$B$ and $C = B ∩ C$
$= {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}$
$A$ and $E = A ∩ E$
$= {(4, 6), (6, 4), (6, 5), (5, 6), (6, 6)}$
$A$ or $F = A ∪ F$
$= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6),$
$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2),$
$(6, 3), (6, 4), (6, 5), (6, 6) }$
$A$ and $F = A ∩ F$
$= {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}$

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Question 852 Marks

A coin is tossed. If it shows tail, we draw a ball from a box which contains $2$ red $3$ black balls; if it shows head, we throw a die. Find the sample space of this experiment.

Answer

The box contains two red balls and three black balls.
Let us denote the red balls as $R_1$ and $R_2$ and the three black balls as $B_1, B_2$ and $B_3.$
The sample space of this experiment is given by $\mathrm{S = \{(T, R_1), (T, R_2), (T, B_1), (T, B_2), (T, B_3), (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)\}.}$

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Question 862 Marks

A coin is tossed twice. If the second throw results in a tail, a die is thrown. Describe the sample space for this experiment.

Answer

When a coin is tossed twice, the possible outcomes are HH, HT, TH, TT.
The second throw results in a head in HH, TH.
The second throw results in a tail in HT, TT.
Now, a dice is thrown.
The possible outcomes on a dice are 1, 2, 3, 4, 5 and 6. The sample space is given by S = { HH, TH, (HT, 1), (HT, 2), (HT, 3), (HT, 4), (HT, 5), (HT, 6), (TT, 1), (TT, 2), (TT, 3) (TT, 4), (TT, 5), (TT 6)}.
$\therefore$The total number of elements of the sample space is 14.

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2 Marks Questions - Page 2 - Maths STD 11 Science Questions - Vidyadip