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MATHS 2 Marks Questions

Probability · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 117 questions.

Questions · Page 2 of 3

2 Marks Questions

Question 512 Marks
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
at least one will fuse after 150 days of use.
Answer
Let X be the number of bulbs that fuse after 150 days.
X follows a binomial distribution with n = 5, p = 0.05 and q = 0.95
Or $\text{p}=\frac{1}{20}$ and $\text{q}=\frac{19}{20}$
$\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{20}\big)^{\text{r}}\big(\frac{19}{20}\big)^{5-\text{r}}$
Probability that that at lesast one will fuse
$=\text{P}(\text{X}=1)+\text{P}(\text{X}=2)+\text{P}(\text{X}=3)+\text{P}(\text{X}=4)+\text{P}(\text{X}=5)$
$=1-\text{P}(\text{X}=0)$
$=1-\Big[\text{ }^5\text{C}_0\big(\frac{1}{20}\big)^0\big(\frac{19}{20}\big)^{5-0}\Big]$
$=1-\Big[\big(\frac{19}{20}\big)^5\Big]$
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Question 522 Marks
A discrete random variable $X$ has the probability distribution given below:
$X:$ $0.5$ $1$ $1.5$ $2$
$P(X):$ $k$ $k^2$ $2k^2$ $k$
Find the value of $k.$
Answer
We know that,
$P(0.5) + P(1) + P(1.5) + P(2) = 1$
$k + k^2+ 2k^2 + k = 1$
$3k^2 + 2k - 1 = 0$
$3k^2 + 3k - k - 1 = 0$
$(3k - 1)(k + 1) = 0$
$\text{k}=\frac{1}{3}$ or $\text{k}=-1$
We know that $0\leq\text{P}(\text{X})\leq1$
$\therefore\ \text{k}=\frac{1}{3}$
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Question 532 Marks
A factory produces bulbs. The probability that one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10. From a single box, find the probability that. 
exactly two bulbs are defective.
Answer
Let getting a defective bulb from a single box is a success.
We have,
p = probability of getting a defective bulb $=\frac{1}{50}$
Also, $\text{q}=1-\text{p}=1-\frac{1}{50}=\frac{49}{50}$
Let X denote the number of success in a sample of 10 trials. Then,
X follows binomial distribution with parameters n = 10 and $\text{p}=\frac{1}{50}$
$\therefore\text{P(X = r})=\text{ }^{10}\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(10-\text{r})}=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{50}\big)^{\text{r}}\big(\frac{49}{50}\big)^{(10-\text{r})}=\frac{\text{ }^{10}\text{C}_{\text{r}}49^{(10-\text{r})}}{50^{10}},$ where r = 0, 1, 2, 3, .....,10
Now,
Required probability = P(exactly two bulbs are defective)
$=\text{P(X}=2)$
$=\frac{\text{ }^{10}\text{C}_249^8}{50^{10}}$
$=\frac{45\times49^8}{50^{10}}$
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Question 542 Marks
A and B are two events such that P (A) ≠ 0. Find P(B|A), if
  1. A is a subset of B
  2. $\text{A}\cap\text{B}=\phi$
Answer
A and B are two events such that P (A) ≠ 0. To find: P(B|A)
  1. $\text{A is a subset of B}\ \Rightarrow\ \text{A}\subset\text{B}$
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})$
$\therefore\ \text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{\text{P}(\text{A})}{\text{P}(\text{A})}=1$
  1. $\text{A}\cap\text{B}=\phi$
$\therefore\ \text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{0}{\text{P}(\text{A})}=0$
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Question 552 Marks
Determine P(E|F) in Exercises. A die is thrown three times,E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.
Answer
Since a dice has six faces. Therefore n (S) = 6 × 6 × 6 = 216
E = (1, 2, 3, 4, 5, 6) × (1, 2, 3, 4, 5, 6) × (4)
F = (6) × (5) × (1, 2, 3, 4, 5, 6) $\Rightarrow$ n(F) = 1 × 1 × 6 = 6
$\text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{6}{216}$
$\therefore\ \ \ \ \ \ \ \text{E}\cap\text{F}=\left(6, 5, 4\right)\ \ \ \ \ \Rightarrow\ \ \ \text{n}\left(\text{E}\cap\text{F}\right)=1$
$\therefore\ \ \ \ \ \ \ \text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\cap\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{1}{216}$
$\text{And}\ \ \ \ \text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\cap\text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{216}$
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Question 562 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 572 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 582 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 592 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 602 Marks
The mean and variance of a binomial distribution are $\frac{4}{3}$ and $\frac{8}{9}$ respectively. Find P (X ≥ 1).
Answer
$=1-\big(\frac{2}{3}\big)^{4}$
$=1-\frac{16}{81}$
$=\frac{65}{81}$
$\text{P(X}\geq1)=\frac{65}{81}$
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Question 612 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 622 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 632 Marks
In a binomial distribution, if n = 20 and q = 0.75, then write its mean.
Answer
$\text{n}=20,\text{q}=0.75$
$\Rightarrow\text{p}=1-\text{q}=0.25$
$\text{Mean = np}=20(0.25)=5$
Thus, $\text{Mean = 5}$
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Question 642 Marks
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable ?
Answer
There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.
Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.
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Question 652 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 662 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 672 Marks
If the probability distribution of a random variable $X$ is given below:
$X = X_i$ $1$ $2$ $3$ $4$
$P(X = X_i)$ $c$ $2c$ $4c$ $4c$
Write the value of $\text{P}(\text{X}\leq2)$
Answer
We know that the sum of probabilities in a probability distribution is always $1.$
Therefore,
$P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1$
$\Rightarrow c + 2c + 4c + 4c =1$
$\Rightarrow 11c = 1$
$\Rightarrow\text{c}=\frac{1}{11}$
Now,
$\text{P}(\text{X}\leq2)=\text{P}(\text{X}=1)+\text{P}(\text{X})=2=\frac{1}{10}+\frac{2}{10}=\frac{3}{10}$
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Question 682 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 692 Marks
A fair coin is tossed 8 times, find the probability of.
at least six heads.
Answer
Probability of getting atleast 6 heads
$=\text{P}(\text{X}\geq6)$
$=\text{P}(\text{X}=6)+\text{P}(\text{X}=7)+\text{P}(\text{X}=8)$
$=\text{ }^8\text{C}_6\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_7\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_8\big(\frac{1}{2}\big)^8$
$=(28+8+1)\times\frac{1}{256}$
$=\frac{37}{256}$
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Question 702 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 712 Marks
If in a binomial distribution mean is 5 and variance is 4, write the number of trials.
Answer
$\text{Mean}=5$ and $\text{Variance}=4$
$\Rightarrow\text{np}=5$ and $\text{npq}=4$
$\Rightarrow\text{q}=0.8$
$\Rightarrow\text{p}=1-\text{q}=0.2$
$\&\text{ np}=\text{n}(0.2)=5$ (given)
$\Rightarrow\text{n}=\frac{5}{0.2}=25$
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Question 722 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 732 Marks
If on an average 9 ships out of 10 arrive safely at ports, find the mean and S.D. of the ships returning safely out of a total of 500 ships.
Answer
Total number of ships (n) = 500
Let X denote the number of ships returning safely to the ports.
$\text{p}=\frac{9}{10}$ and $\text{q}=1-\text{p}=\frac{1}{10}$
$\text{Mean = np = 450}$ and $\text{Variance = npq = 45}$
$\text{Mean = 450}$
$\text{Standard deviation}=\sqrt{45}=6.71$
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Question 742 Marks
Six coins are tossed simultaneously. Find the probability of getting.
at least one head.
Answer
Let p represents the probability of getting head in a toss of fair coin, so
$\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$ [Since p + q = 1]
$\text{q}=\frac{1}{2}$
Let X denote the random variable representing the number heads in 6 tosses of coin. probability of getting r sixes in n tosses of a fair coin is given by,
$\text{P(X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$3
$=\text{ }^6\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}}\dots(1)$
P(getting at least 1 head) $=\text{P(X}\geq1)$
$=1-\text{P(X}=0)$
$=1-\frac{1}{64}$
$=\frac{63}{64}$
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Question 752 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 762 Marks
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
none.
Answer
Let X be the number of bulbs fuse after 150 days.
X follows a binomial disribution with n = 5, p = 0.05 and q = 0.95
Or $\text{p}=\frac{1}{20}$ and $\text{q}=\frac{19}{20}$
$\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{20}\big)^{\text{r}}\big(\frac{19}{20}\big)^{5-\text{r}}$
Probability (nine will fuse after 150 days of use) = P(X = 0)
$=\text{ }^5\text{C}_0\big(\frac{1}{20}\big)^0\big(\frac{19}{20}\big)^{5-0}$
$=\big(\frac{19}{20}\big)^5$
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Question 772 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 782 Marks
A random variable has the following probability distribution:
$X = X_i$ $1$ $2$ $3$ $4$
$P(X = X_i)$ $k$ $2k$ $3k$ $4k$
Write the value of $\text{P}(\text{X}\geq3).$
Answer
Here,
$X = X_i$ $1$ $2$ $3$ $4$
$P(X = X_i)$ $k$ $2k$ $3k$ $4k$
Since, $\sum\text{P}(\text{X})=1$
$\Rightarrow P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1$
$\Rightarrow k + 2k + 3k + 4k =1$
$\Rightarrow 10k = 1$
$\Rightarrow\text{k}=\frac{1}{10}$ $\text{P}(\text{X}\geq3)$
$= P(X = 3) + P(X = 4)$
$= 3k + 4k$
$= 7k =\frac{1}{10}$ $\text{P}(\text{X}\geq3)=\frac{7}{10}$
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Question 792 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 802 Marks
Suppose that a radio tube inserted into a certain type of set has probability $0.2$ of functioning more than $500$ hours. If we test $4$ tubes at random what is the probability that exactly three of these tubes function for more than $500$ hours?
Answer
Let $X$ denote the number of tubes that function for more than $500$ hours.
Then, $X$ follows a binomial distribution with $n = 4.$
Let $p$ be the probability that the tubes function more than $500$ hours
Here, $p = 0.2, q = 0.8$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^4\text{C}_{\text{r}}(0.2)^{\text{r}}(0.8)^{4-\text{r}},\text{r}=0,1,2,3,4$
Therefore, required probability $= P(X = 3)$
$= 4(0.2)^{3 }(0.8)$
$= 0.0256$
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Question 812 Marks
A random variable X has the following probability distribution:
Values of X: -2 -1 0 1 2 3
P(X) 0.1 k 0.2 2k 0.3 k
Find the value of k.
Answer
Here,
Values of X: -2 -1 0 1 2 3
P(X) 0.1 k 0.2 2k 0.3 k
We know that,
P(-2) + P(-1) + P(0) + P(1) + P(2) + P(3) = 1
⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1
⇒ 4k + 0.6 = 1
⇒ 4k = 1 - 0.6
⇒ 4k = 0.4
$\Rightarrow\text{k}=\frac{0.4}{4}$
$\Rightarrow\text{k}=\frac{1}{10}$
$\Rightarrow\text{k}=0.1$
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Question 822 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 832 Marks
Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that
all the five cards are spades?
Answer
Let P denote the probability of geting one spade out of a deck of 52 cards, So
$\text{p}=\frac{13}{52}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
$\text{q}=\frac{3}{4}$
let X denote the radom variable of number of spades out of 5 cards. probability of getting r spades out of n cards is given by
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^5\text{c}_\text{r}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{5-\text{r}}\dots(1)$
Probability of getting all five spads
$\text{P}(\text{X}=5)$
$=\text{ }^5\text{c}_5\big(\frac{1}{4}\big)^5\big(\frac{3}{4}\big)^5-5$
$=\frac{1}{1024}$
Probability of getting 5 spades $=\frac{1}{1024}$
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Question 842 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 852 Marks
A factory produces bulbs. The probability that one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10. From a single box, find the probability that.
more than 8 bulbs work properly.
Answer
Let getting a defective bulb from a single box is a success.
We have,
p = probability of getting a defective bulb $=\frac{1}{50}$
Also, $\text{q}=1-\text{p}=1-\frac{1}{50}=\frac{49}{50}$
Let X denote the number of success in a sample of 10 trials. Then,
X follows binomial distribution with parameters n = 10 and $\text{p}=\frac{1}{50}$
$\therefore\text{P(X = r})=\text{ }^{10}\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(10-\text{r})}=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{50}\big)^{\text{r}}\big(\frac{49}{50}\big)^{(10-\text{r})}=\frac{\text{ }^{10}\text{C}_{\text{r}}49^{(10-\text{r})}}{50^{10}},$ where r = 0, 1, 2, 3, .....,10
Now,
Required probility = P(more than 8 bulbs work properly)
= P(atmost one bulb is defective)
$=\text{P(X}\leq0)$
$=\text{P(X}=0)+\text{P(X}=1)$
$=\frac{\text{ }^{10}\text{C}_049^{10}}{50^{10}}+\frac{\text{ }^{10}\text{C}_149^9}{}$
$=\frac{49^{10}}{50^{10}}+\frac{10\times49^{9}}{50^{10}}$
$=\frac{49^6}{50^{10}}(49+10)$
$=\frac{59(49^9)}{(50^{10})}$
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Question 862 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 872 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 882 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 892 Marks
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
E: ‘the card drawn is black’
F: ‘the card drawn is a king’
Answer
In a deck of 52 cards, 26 cards are black and 4 cards are kings.
$\therefore$ P(E) = P(the card drawn is black) $=\frac{26}{52}=\frac{1}{2}$
$\therefore$ P(F) = P(the card drawn is a king) $=\frac{4}{52}=\frac{1}{13}$
In the pack of 52 cards, 2 cards are black as well as kings.
$\therefore$ P (EF) = P(the card drawn is a black king) $=\frac{2}{52}=\frac{1}{26}$
$\text{P}(\text{E})\times\text{P}(\text{F})=\frac{1}{2}\cdot\frac{1}{13}=\frac{1}{26}=\text{P}(\text{EF})$
Therefore, the given events E and F are independent.
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Question 902 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 912 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 922 Marks
The mean of a binomial distribution is $10$ and its standard deviation is $2;$ write the value of $q.$
Answer
Mean of the binomial distribution, i.e. $\text{np}=10$
Variance $= ($Standard deviation$)^2,$ i.e. $\text{npq}=4$
$\therefore\text{q}=\frac{\text{Variance}}{\text{Mean}}=0.4$
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Question 932 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 942 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;
$P (A_W$ and $B_W) = P(A_W) P(B_W)$
$=\frac{4}{6}\times\frac{3}{8}=\frac{1}{4}$
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Question 952 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 962 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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Question 972 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 982 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 992 Marks
In a group of 200 items, if the probability of getting a defective item is 0.2, write the mean of the distribution.
Answer
Let p be the probability of defective item, so
$\text{p}=0.2$
$\text{p}=\frac{1}{5}$
$\text{q}=1-\frac{1}{5}$ [Since p+ q =1]
$\text{q}=\frac{4}{5}$
Given, $\text{n}=200$
$\text{Mean = np}$
$=200\big(\frac{1}{5}\big)$
$=40$
$\text{Mean}=40$
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Question 1002 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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2 Marks Questions - Page 2 - MATHS STD 12 Science Questions - Vidyadip