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MATHS 2 Marks Questions

Probability · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 65 questions.

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Question 12 Marks
In a competition A, B and C are participating. The probability that A wins is twice that of B, the probability that B wins is twice that of C. Find the probability that A losses.
Answer
Let P(A wins) = x
So, P(B wins) = 2x
P(A wins) = 2P(B wins)
= 2(2x)
P(A wins) = 4X
P(A wins) + P(B wins) + P(C wins) = 1
⇒ 4x + 2x+ x = 1
⇒ 7x = 1
$\Rightarrow\ \text{x}=\frac{1}{7}$
$\text{P(A wins)}=4\text{x}$
$=\frac{4}{7}$
$\text{P(A losses)}=1-\text{P(A wins)}$
$=1-\frac{4}{7}$
$=\frac{3}{7}$
Required probability $=\frac{3}{7}$
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Question 22 Marks
If A and B are two independent events such that P(A) = 0.3 and $=0.8\text{P}(\text{A}\cap\overline{\text{B}})$ Find P(B).
Answer
A and B are two independent events
$\therefore\ \text{P}(\text{A}\cap\overline{\text{B}})=\text{P(A)}+\text{P}(\overline{\text{B}})-\text{P}(\text{A}\cap\overline{\text{B}})$
$\Rightarrow\ 0.8=0.3+[1-\text{P(B)}]-\text{P(A)}\text{P}(\overline{\text{B}})$
$\Rightarrow 0.5=1-\text{P(B)}-0.3[1-\text{P(B)}]$
$\Rightarrow 0.5=1-\text{P(B)}=0.3+0.3\text{P(B)}$
$\Rightarrow\ 0.5= 0.7-\text{P(B)}[1-0.3]$
$\Rightarrow\ 0.7\text{P(B)} = 0.2$
$\Rightarrow\ \text{P(B)}=\frac{0.2}{0.7}=\frac{2}{7}$
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Question 32 Marks
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Blue followed by red.
Answer
Bag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Blue followed by red)
$=\text{P}(\text{B}\cap\text{R})$
$=\text{P}(\text{B})\times\text{P}(\text{R})$
$=\frac{3}{8}\times\frac{5}{8}$
$=\frac{15}{64}$
Required probability $=\frac{15}{64}$
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Question 42 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
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
Answer
Given,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11},\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\frac{4}{11}}{\frac{6}{11}}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{2}{3}$
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Question 52 Marks
6 boys and 6 girls sit in a row at random. Find the probability that all the girls sit together.
Answer
Here, 6 bhoys and 6 girls can be arranged in a line in 12! ways.
Total possible outcomes = 12!
Consider 6 girls as a single element X.
Now, 6 boys and X can be arranged in aline in 7! ways and girls can be arranged in 6! ways among them.
P(all girls are together) $=\frac{7!\times6!}{12!}$
$=\frac{7\times6\times5\times4\times3\times2\times1\times6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}$
$=\frac{1}{11\times12}$
$=\frac{1}{132}$
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Question 62 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
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
Answer
Given,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11},\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{\frac{4}{11}}{\frac{5}{11}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{4}{5}$
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Question 72 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,
Only one 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(Only one of them will be selected)
$=\text{P}\Big[(\text{H}\cap\overline{\text{W}})\cup(\overline{\text{H}}\cap\text{W})\Big]$
$=\text{P}(\text{H}\cap\overline{\text{W}})+\text{P}(\overline{\text{H}}\cap\text{W})$
$=\text{P(H)}\text{ P}(\overline{\text{W}})+\text{P}(\overline{\text{H}})\text{ P(W)}$
$=\text{P(H)}[1-\text{P(W)}]+[1-\text{P(H)}]\text{ P(W)}$
$=\frac{1}{7}\Big[1-\frac{1}{5}\Big]+\Big[1-\frac{1}{7}\Big]\frac{1}{5}$
$=\frac{1}{7}\times\frac{4}{5}+\frac{6}{7}\times\frac{1}{5}$
$=\frac{10}{35}$
$=\frac{2}{7}$
Required probability $=\frac{2}{7}$
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Question 82 Marks
Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
Both are kings.
Answer
In a deck of 52 cards. there are 4 kings. Two cards are drawn without replacement,
A = First card is king
B = Second card is king
P (Both drawn cards are king)
$=\text{P}(\text{A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}$
$=\frac{1}{221}$
Required Probability $=\frac{1}{221}$
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Question 92 Marks
If A and B are two events write the expression for the probability of occurrence of exactly one of two events.
Answer
P(exaxtly one of 2 events) $=\text{P}(\text{A}\cup\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}+\text{P(B)}-2\text{P}(\text{A}\cap\text{B})$
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Question 102 Marks
An unbiased die with face marked 1, 2, 3, 4, 5, 6 is rolled four times. Out of 4 face values obtained, find the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5.
Answer
Die is rolled 4 times
n(S) = Number of elements in samplw space
n(S) = 6 × 6 × 6 × 6
Given,
Number obtained on face are not less than 2 and greater than 5.
E = Obtaining 2, 3, 4, 5 on die in four throw
n(E) = 4 × 4 × 4 × 4
$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}$
$=\frac{4\times4\times4\times4}{6\times6\times6\times6}$
$=\frac{16}{81}$
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Question 112 Marks
If two events A and B are such that $\text{P}(\overline{\text{A}})=0.3,\text{P(B)}=0.4$ and $\text{P}(\text{A}\cap\overline{\text{B}})=0.5$ find $\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big).$
Answer
According to Baye's Theorem
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=\frac{\text{P}(\text{B}\cap(\overline{\text{A}}\cap\overline{\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\text{B}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
Now $\overline{\text{B}}\cap\text{B}=\cup=\phi$
So, $\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))=\phi$
$\therefore\ \text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=0$
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Question 122 Marks
When three dice are thrown, write the probability of getting 4 or 5 on each of the dice simultaneously.
Answer
There dice are thrown
Given, 4 or 5 on each of the dice simultaneously
= {(4, 4, 4), (4, 4, 5), (4, 5, 4), (4, 5, 5), (5, 4, 4), (5, 4, 5), (5, 5, 4), (5, 5, 5)}
n(E) = 8
n(S) = 216
$\text{P(E)}=\frac{8}{216}$
$=\frac{1}{27}$
Required probability $=\frac{1}{27}$
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Question 132 Marks
Let $A$ and $B$ be two independent events such that $P(A) = p_1$ and $P(B) = p_2.$ Describe in words the events whose probabilities are:
$(1 - p_1)p_2.$
Answer
As, $(1 - p_1)p_2$
$=[1-\text{P(A)}]\times\text{P(B)}=\text{P}(\overline{\text{A}})\times\text{P(B)}$
And, $A$ and $B$ are independent events.
i.e., $\text{P}(\overline{\text{A}})\times\text{P}(\text{B})=\text{P}(\text{A}\cap\text{B})$
So, $\text{P}(\text{A}\cap\text{B})=\text{p}_1\text{p}_2$
Hence, $(1 - p_1)p_2 = P (A$ does not occur, but $B$ occurs$).$
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Question 142 Marks
An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
2 red balls.
Answer
An urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 red balls)
$=\text{P}(\text{R}_1)\times\text{P}(\text{R}_2)$
$=\frac{7}{11}\times\frac{7}{11}$
$=\frac{49}{121}$
Required probability $=\frac{49}{121}$
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Question 152 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,
None 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(None of them selected)
$=(\overline{\text{H}}\cap\overline{\text{W}})$
$=\text{P}(\overline{\text{H}})\text{ P}(\overline{\text{W}})$
$=(1-\text{P(H)})(1-\text{P(W)})$
$=\Big(1-\frac{1}{7}\Big)\Big(1-\frac{1}{5}\Big)$
$=\frac{6}{7}\times\frac{4}{5}$
$=\frac{24}{35}$
Required probability $=\frac{24}{35}$
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Question 162 Marks
Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among 100 students, what is the probability that:
You both enter the different sections?
Answer
P (Both enter different section)
= 1 - P(Both enter same section)
$=1-\frac{17}{33}$
$=\frac{16}{33}$
P(Both eneter different section) $=\frac{16}{33}$
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Question 172 Marks
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Of the same colour.
Answer
Bag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Of the same colour)
$=\text{P}\big((\text{R}_1\cap\text{R}_2)\cup(\text{B}_1\cap\text{B}_2)\big)$
$=\text{P}(\text{R}_1)\times\text{P}(\text{R}_2)+\text{P}(\text{B}_1)\times\text{P}(\text{B}_2)$
$=\frac{5}{8}\times\frac{5}{8}+\frac{3}{8}\times\frac{3}{8}$
$=\frac{25+9}{64}$
$=\frac{34}{64}$
$=\frac{17}{32}$
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Question 182 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cap\text{B})$
Answer
Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cap\text{B})=\text{P(A) }\text{P(B)}$
[Since, A and B are indenpendent events]
$=0.3\times0.6$
$\text{P}(\text{A}\cap\text{B})=0.18$
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Question 192 Marks
Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
The first is a heart and second is red.
Answer
There are 13 heart and 26 red cards
Hearts are also red.
A = first card is heart.
B = second card is red.
P (First card is heart and second is red)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{13}{52}\times\frac{25}{51}$
$=\frac{25}{204}$
Required probability $=\frac{25}{204}$
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Question 202 Marks
An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
One red and one blue ball.
Answer
An urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 blue balls)
$=\text{P}\big((\text{R}\cap\text{B})\cup(\text{B}\cap\text{R})\big)$
$=\text{P(R)}\times\text{P(B)}+\text{P(B)}\times\text{P(R)}$
$=\frac{7}{11}\times\frac{4}{11}+\frac{4}{11}\times\frac{7}{11}$
$=\frac{28+28}{121}$
$=\frac{56}{121}$
Required probability $=\frac{56}{121}$
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Question 212 Marks
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Blue and red in any order.
Answer
Bag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Blue and red in any order)
$=\text{P}\big((\text{B}\cap\text{R})\cup(\text{R}\cap\text{B})\big)$
$=\text{P}(\text{B}\cap\text{R})+\text{P}(\text{R}\cap\text{B})$
$=\text{P}(\text{B})\times\text{P}(\text{R})+\text{P}(\text{R})\times\text{P}(\text{B})$
$=\frac{3}{8}\times\frac{5}{8}+\frac{5}{8}\times\frac{3}{8}$
$=\frac{30}{64}$
$=\frac{15}{32}$
Required probability $=\frac{15}{32}$
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Question 222 Marks
If $\text{P(A)}=0.3,\text{P(B)}=0.6,\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5,$ find $\text{P}(\text{A}\cup\text{B}).$
Answer
Given,
$\text{P(A)}=0.3,\text{P(B)}=0.6,\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5,$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$0.5=\frac{\text{P}(\text{A}\cap\text{B})}{0.3}$
$\text{P}(\text{A}\cap\text{B})=0.5\times0.3$
$\text{P}(\text{A}\cap\text{B})=0.15$
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.3+0.6+0.15$
$=0.75$
$\text{P}(\text{A}\cup\text{B})=0.75$
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Question 232 Marks
Three digit numbers are formed with the digits 0, 2, 4, 6 and 8. Write the probability of forming a three digit number with the same digits.
Answer
Total 3-digit numbers that can be made out of 0, 2, 4, 5 and 8 = 4 × 5 × 5 (hundreds place cannot be filled with 0)
= 100
But 222, 444, 666 and 888 are four numbers, which have the same digits at all places.
P(3-digit number having same digits at all places) $=\frac{4}{100}$
$=\frac{1}{25}$
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Question 242 Marks
An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
2 Blue balls.
Answer
An urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 blue balls)
$=\text{P}(\text{B}_1\cap\text{B}_2)$
$=\text{P}(\text{B}_1)\times\text{P}(\text{B}_2)$
$=\frac{4}{11}\times\frac{4}{11}$
$=\frac{16}{121}$
Required probability $=\frac{16}{121}$
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Question 252 Marks
A bag contains $4$ white balls and $2$ black balls. Another contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag, find the probability that,
Both are black.
Answer
Bag $A$ has $4$ white balls and $2$ black balls;
Bag $B$ has $3$ white balls and $5$ black balls;
$P (A_B$ and $B_B) = P(A_B) P(B_B)$
$=\frac{2}{6}\times\frac{5}{8}=\frac{5}{24}$
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Question 262 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.
Both 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(Both get selected) = P(A) × P(B)
$=\frac{1}{3}\times\frac{1}{5}$
$=\frac{1}{15}$
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Question 272 Marks
A bag contains $4$ white balls and $2$ black balls. Another contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag, find the probability that,
One is while and one is black.
Answer
Bag $A$ has $4$ white balls and $2$ black balls;
Bag $B$ has $3$ white balls and $5$ black balls;
$P (A_W$ and $B_B$ or $A_B$ and $B_W)$
$= P(A_W) P(B_B) + P(A_B) P(B_W)$
$=\frac{4}{6}\times\frac{5}{8}+\frac{2}{6}\times\frac{3}{8}$
$=\frac{20}{48}+\frac{6}{48}$
$=\frac{26}{48}=\frac{13}{24}$
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Question 282 Marks
A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, find the probability that these two balls are of the same colour.
Answer
Given:
Bag 1 = (3W + 6B) balls
Bag 2 = (5B + 4W) balls
P(balls of same colour are drawn) = P(both black) + P(both white)
$=\frac{6}{9}\times\frac{5}{9}+\frac{3}{9}\times\frac{4}{9}$
$=\frac{30}{81}+\frac{12}{81}$
$=\frac{42}{81}$
$=\frac{14}{27}$
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Question 292 Marks
Three cards are drawn with replacement from a well shuffled pack of cards. Find the probability that the cards drawn are king, queen and jack.
Answer
$\text{P (king)}=\frac{4}{52}$
$\text{P (queen)}=\frac{4}{52}$
$\text{P (jack)}=\frac{4}{52}$
These cards can be drawn in $^3P_3$ ways.
$\text{P (king, queen, and jack)}=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}\times {^{3}}\text{P}_3$
$=\frac{3!}{2197}$
$=\frac{6}{2197}$
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Question 302 Marks
A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Write the probability that the number is divisible by 5.
Answer
To be divisibel by 5 ones place sholud be 5
There are 3 place remaining which can be filled in 3! = 6 ways
So, 6 number can be formed out of 1, 2, 3 and 5, which are divisible by 5.
Total 4 - digit numbers = 4! = 24
P (4-digit number divisible by 5) $=\frac{6}{24}$
$=\frac{1}{4}$
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Question 312 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cup\text{B})$
Answer
Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cup\text{B})=\text{P(A)} + \text{P(B)} - \text{P}(\text{A}\cap\text{B})$
$=0.3+0.6-0.18$
$\text{P}(\text{A}\cup\text{B})=0.72$
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Question 322 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 both the cards are of the same suit?
Answer
P(both the cards are of same suit) = P(both the cards are of diamond) + P(both the cards are of spade) + P(both the cards are of club) + P(both the cards are of hear)
$=\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}$
$=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}$
$=\frac{4}{16}$
$=\frac{1}{4}$
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Question 332 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,
Both 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(Boht of them selected)
$=\text{P}(\text{F}\cap\text{J})$
$=\text{P}(\text{F})\text{P}(\text{J})$
$=\frac{1}{7}\times\frac{1}{5}$
$=\frac{1}{35}$
Required probability $=\frac{1}{35}$
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Question 342 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cap\overline{\text{B}})$
Answer
Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cap\overline{\text{B}})=\text{P(A)}-\text{P}(\text{A}\cap\text{B})$
$=0.3 - 0.18$
$\text{P}(\text{A}\cap\overline{\text{B}})=0.12$
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Question 352 Marks
If $\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13},$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big).$
Answer
Given:
$\text{P(A)}=\frac{7}{13}$
$\text{P(B)}=\frac{9}{13}$
$\text{P}(\text{A}\cap\text{B})=\frac{4}{13}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac{4}{9}$
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Question 362 Marks
Two cards are drawn from a well shuffled pack of 52 cards, one after another without replacement. Find the probability that one of these is red card and the other a black card?
Answer
P(One red and one black) = P(first red and second black) + P(first black and second red)
$=\frac{26}{5}\times\frac{26}{51}+\frac{26}{52}\times\frac{26}{51}$
[Without replacement]
$=\frac{13}{51}+\frac{13}{51}$
$=\frac{26}{51}$
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Question 372 Marks
A ordinary cube has four plane faces, one face marked $2$ and another face marked $3,$ find the probability of getting a total of $7$ in $5$ throws.
Answer
A cube has total $6$ faces.
Total possible outcomes in $5$ throws $= 6 \times 6\times 6 \times 6 \times 6 = (6)^5$
The only way of getting $7$ is by getting two $2s$ and one $3.$
Total possible ways $=\frac{\text{P}^5_3}{2!}$
$=\frac{5\times4\times3\times2\times1}{2\times1\times2\times1}$
$=30$
Now,
$P($getting $7$ in $5$ throws $)=\frac{30}{6^5}=\frac{5}{6^4}$
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Question 382 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.
Only one 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 (One of them gets selected) $=\text{P}(\overline{\text{A}})\text{ P(B)}+\text{P}(\overline{\text{B}})\text{ P(A)}$
$=\text{P(B)}[1-\text{P(A)}]+\text{P(A)}[1-\text{P(B)}]$
$=\frac{1}{5}\Big(1-\frac{1}{3}\Big)+\frac{1}{3}\Big(1-\frac{1}{5}\Big)$
$=\frac{2}{15}+\frac{4}{15}$
$=\frac{6}{15}=\frac{2}{5}$
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Question 392 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
First ball is black and second is red.
Answer
Given,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (First ball is black and second is red)
$=\text{P}(\text{B}\cap\text{R})$
$=\text{P(B)}\text{ P(R)}$
$=\frac{10}{18}\times\frac{8}{18}$
$=\frac{20}{81}$
Required probability $=\frac{20}{81}$
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Question 402 Marks
Let $A$ and $B$ be two independent events such that $P(A) = p_1$ and $P(B) = p_2.$ Describe in words the events whose probabilities are:
$p_1p_{2}.$
Answer
As, $p_1p_2 = P(A) \times P(B)$
And, $A$ and $B$ are independent events.
i.e., $\text{P(A)}\times\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, $\text{P}(\text{A}\cap\text{B})=\text{P}_1\text{P}_2$
Hence, $p_1p_2 = P(A$ and $B$ occur$).$
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Question 412 Marks
The probability of student A passing an examination is $\frac{2}{9}$ and of student B passing is $\frac{5}{9}$. Assuming the two events: 'A passes', 'B passes' as independent, find the probability of:
Only A passing the examination.
Answer
Given,
The probability of A passing exam $=\frac{2}{9}$
The probability of B passing exam $=\frac{5}{9}$
$\Rightarrow\ \text{P(A)}=\frac{2}{9},\text{P(B)}=\frac{5}{9}$
P (Only A passing the exam)
$=\text{P}(\text{A}\cap\overline{\text{B}})$
$=\text{P(A)}\text{P(B)}$
$=\text{P(A)}[1-\text{P(B)}]$
$=\frac{2}{9}\Big(1-\frac{5}{9}\Big)$
$=\frac{2}{9}\Big(\frac{4}{9}\Big)$
$=\frac{8}{81}$
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Question 422 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
One of them is black and other is red.
Answer
Given,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (one of them red and other black)
$=\text{P}\big((\text{B}\cap\text{R})\cup(\text{R}\cap\text{B})\big)$
$=\text{P}(\text{B}\cap\text{R})+\text{P}(\text{R}\cap\text{B})$
$=\text{P(B) }\text{P(R)}+\text{P(R) }\text{P(B)}$
$=\frac{10}{18}\times\frac{8}{18}+\frac{8}{18}\times\frac{10}{18}$
$=\frac{20+20}{81}$
$=\frac{40}{81}$
Required probability $=\frac{40}{81}$
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Question 432 Marks
A bag contains $4$ white balls and $2$ black balls. Another contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag, find the probability that, Both are white.
Answer
Bag $A$ has $4$ white balls and $2$ black balls;
Bag $B$ has $3$ white balls and $5$ black balls;
$\ce{P (A_W}$ and $\ce{B_W) = P(A_W) P(B_W)}$
$=\frac{4}{6}\times\frac{3}{8}=\frac{1}{4}$
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Question 442 Marks
Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that,
  1. The youngest is a girl.
  2. At least one is a girl?
Answer
Let a and g rapresent the boyy and the girl child respectively. if a family has two children, the samplw space will be S = {(b, b), (b, g), (g, b), (g, g)} Let a be the event that both children are girls. $\therefore\ \text{A}=\big\{(\text{g},\text{g})\big\}$
  1. Let B be the event that the youngest child is a girl.
$\therefore\text{B}=\big[(\text{b},\text{g}),(\text{g},\text{g})\big]$
$\Rightarrow\ \text{A}\cap\text{B}=\big\{(\text{g},\text{g})\big\}$
$\therefore\text{P(B)}=\frac{2}{4}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
The conditional probability that both are girls, given that the youngest child is a girl, is given by p (A | B).
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$
Therefore, the required probability is $\frac{1}{2}$.
  1. Let C be the event that at least one child is a girl.
$\therefore\text{C}=\big\{(\text{b},\text{g}),(\text{g},\text{b}),(\text{g},\text{g})\big\}$
$\Rightarrow\text{A}\cap\text{C}=\big\{\text{g},\text{g}\big\}$
$\Rightarrow\text{P(C)}=\frac{3}{4}$
$\text{P}(\text{A}\cap\text{C})=\frac{1}{4}$
The conditional probability that both are girls, given that at least one child a girl, is given by P(A|C).
$\therefore \text{P}(\text{A}|\text{C})=\frac{\text{P}(\text{A}\cap\text{C})}{\text{P}(\text{C})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
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Question 452 Marks
X is taking up subjects - Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets,
Grade A in no subject.
Answer
Given,
Probability of getting A grade in mathematics (m) = 0.2
⇒ P(m) = 0.2
Probability of getting A grade in physics (p) = 0.3
⇒ P(p) = 0.3
Probability of getting A grade in chemistry (P) = 0.5
⇒ P(C) = 0.5
P(Getting A in no subject)
$=\text{P}(\overline{\text{m}}\cap\overline{\text{p}}\cap\overline{\text{c}}\big)$
$=\text{P}(\overline{\text{m}})+\text{P}(\overline{\text{p}})+\text{P}(\overline{\text{c}}\big)$
$=\big(1-\text{P(m)}\big)\big(1-\text{P(p)}\big)\big(1-\text{P(c)}\big)$
$=(1-0.2)(1-0.3)(1-0.5)$
$=(0.8)(0.7)(0.5)$
$=0.28$
Required probability = 0.28
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Question 462 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
Answer
Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.18}{0.6}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.3$
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Question 472 Marks
X is taking up subjects - Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets,
Grade A in all subjects
Answer
Given,
Probability of getting A grade in mathematics (m) = 0.2
⇒ P(m) = 0.2
Probability of getting A grade in physics (p) = 0.3
⇒ P(p) = 0.3
Probability of getting A grade in chemistry (P) = 0.5
⇒ P(C) = 0.5
P(Getting A grade in all subjects)
$=\text{P}(\text{m}\cap\text{P}\cap\text{C})$
= P(m) + P(p) + P(c)
= 0.2 × 0.3 × 0.5
= 0.3
Required probability = 0.03
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Question 482 Marks
If A and B are independent events, then write expression for P(exactly one of A, B occurs).
Answer
As, A and B are independent events.
So, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}\ .....\text{(i)}$
P(exactly one of A, B occurs) = P(only A) + P(only B)
$=\big[\text{P(A)}-\text{P}(\text{A}\cap\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$=\big[\text{P(A)}-\text{P}(\text{A})\times\text{P}(\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A})\times\text{P}(\text{B})\big]$ [Using (i)]
$=\text{P(A)}\times\big[1-\text{P(B)}\big]+\text{P(B)}\times\big[1-\text{P(A)}\big]$
$=\text{P(A)}\times\text{P}(\overline{\text{B}})+\text{P(B)}\times\text{P}(\overline{\text{A}})$
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Question 492 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.
At least one 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 (At least one of them gets selected) $=\text{P}(\text{A}\cup\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P(A)}\times\text{P(B)}$
$=\frac{1}{3}+\frac{1}{5}-\frac{1}{3}\times\frac{1}{5}$
$=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}$
$=\frac{7}{15}$
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Question 502 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 $20$
$n(S) = ^{20}C_3$
$E =$ 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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2 Marks Questions - MATHS STD 12 Science Questions - Vidyadip