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Binomial Distribution · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 32 questions.

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32 questions · timed · auto-graded

Question 12 Marks
Six coins are tossed simultaneously. Find the probability of getting.
3 heads.
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
$=\text{ }^6\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}}\dots(1)$
Probability of getting 3 heads
$=\text{P(X}=3)$
$=\text{ }^6\text{c}_3\big(\frac{1}{2}\big)^3\big(\frac{1}{2}\big)^{6-3}$
$=\frac{6\times5\times4}{3\times2}\big(\frac{1}{2}\big)^3\big(\frac{1}{2}\big)^3$
$=\frac{20}{64}$
Probability of getting 3 heads $=\frac{20}{64}=\frac{5}{16}$
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Question 22 Marks
A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
any two are white?
Answer
Let P be the probability of getting 1 white ball out of 7 red, 5 white and 8 black balls. so
$\text{p}=\frac{5}{20}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
Let X denote the random variable of number of selecting white ball with replacement out of 4 balls. probability of getting r white balls out of n balls is given by
$\text{P}(\text{x = r})=\text{ }^\text{n}\text{c}_\text{r}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^4\text{c}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{4-\text{r}}\dots(1)$
Prop that any two are white $=\text{P}(\text{X}=2)=\text{ }^4\text{c}_2\big(\frac{1}{4}\big)^2\big(\frac{3}{4}\big)^{4-2}=\frac{54}{256}=\frac{27}{128}$
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Question 32 Marks
An unbiased coin is tossed 4 times. Find the mean and variance of the number of heads obtained.
Answer
We have,
p = probability of getting a head in a toss $=\frac{1}{2},$
q = probability of getting a tail in a toss $=\frac{1}{2}$
Let X denote a success of getting a head in a toss, Then,
X follows binomial distribution with parmeters $\text{n}=4$ and $\text{p}=\frac{1}{2}$
$\therefore\text{Mean, E(X) = np}=4\times\frac{1}{2}=2$
Also, $\text{variance, Var(X) = npq}=4\times\frac{1}{2}\times\frac{1}{2}=1$
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Question 42 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 52 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.
not more than one.
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 (not more than 1 will fuse after 150 days of use) = P(X ≤ 1)
$=\text{P}(\text{X}=0)+\text{P}(\text{X}=1)$
$=\big(\frac{19}{20}\big)^5+5\text{C}_1\big(\frac{1}{20}\big)^1\big(\frac{19}{20}\big)^{5-1}$
$=\big(\frac{19}{20}\big)^4\Big\{\frac{19}{20}+\frac{5}{20}\Big\}$
$=\frac{6}{5}\big(\frac{19}{20}\big)^4$
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Question 62 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.
none of the bulbs is 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 probility = P(none of the bulb is defective)
$=\text{P(X}=0)$
$=\frac{\text{ }^{10}\text{C}_049^{10}}{50^{10}}$
$=\frac{49^{10}}{50^{10}}$
$=\big(\frac{49}{50}\big)^{10}$
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Question 72 Marks
Suppose X has a binomial distribution with n = 6 and p = $\frac{1}{2}.$ Show that X = 3 is the most likely outcome.
Answer
we have $\text{ n}=6$ and $\text{p}=\frac{1}{2}$
$\therefore\text{q}=1-\text{p}=\frac{1}{2}$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}},\text{r}=0,1,2,3,4,5,6$
$=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{6}$
$\text{P(X = r})=\frac{\text{ }^6\text{C}_{\text{r}}}{2^6}$
By substituting r = 0, 1, 2, 3, 4, 5 and 6, we get the following
distribution for X.
$\text{x}$
 $0$
 $1$
 $2$
 $3$
 $4$
 $5$
 $6$
$\text{P(X)}$
 $\frac{1}{64}$
 $\frac{6}{64}$
 $\frac{15}{64}$
 $\frac{20}{64}$
 $\frac{15}{64}$
 $\frac{6}{64}$
 $\frac{1}{64}$
 Comparing the probabilites, we get that X = 3 is the most likely outcome
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Question 82 Marks
Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that
only 3 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
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^5\text{c}_\text{r}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{5-\text{r}}\dots(1)$
P(only 3 crads are spades) = P(x = 3)
$=\text{ }^5\text{C}_3\big(\frac{1}{4}\big)^3\big(\frac{3}{4}\big)^2$
$=\frac{1}{1024}(90)$
$=\frac{45}{512}$
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Question 92 Marks
A die is thrown three times. Let X be 'the number of twos seen'. Find the expectation of X.
Answer
Let p be the probability of getting 2 when a dice is thrown.
Then $\text{p}=\frac{1}{6}$
Clearly X follows binomial distribution with $\text{n}=3,\text{p}=\frac{1}{6}.$
$\therefore$ Expectation $=\text{E(X) = np}=3\times\frac{1}{6}=\frac{1}{2}$
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Question 102 Marks
A fair coin is tossed 8 times, find the probability of.
at most six heads.
Answer
Probability of getting at most 6 heads
$=\text{P}(\text{X}\leq6)$
$=1-\big[\text{P}(\text{X}=7)+\text{P}(\text{X}=8)\big]$
$=1-\big[\text{ }^8\text{C}_7\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_8\big(\frac{1}{2}\big)^8\big]$
$=1-\big(\frac{8}{256}+\frac{1}{256}\big)$
$=1-\frac{9}{256}$
$=\frac{247}{256}$
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Question 112 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 122 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 132 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 142 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 152 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 162 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.
more than one.
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 more than one will fuse
$=\text{P}(\text{X}=2)+\text{P}(\text{X}=3)+\text{P}(\text{X}=4)+\text{P}(\text{X}=5)$
$=1-\big[\text{P}(\text{X}=0)+\text{P}(\text{X}=1)\big]$
$=1-\Big[\frac{6}{5}\big(\frac{19}{20}\big)^4\Big]$
Probability that more than one will fuse $=1-\Big[\frac{6}{5}\big(\frac{19}{20}\big)^4\Big]$
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Question 172 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 182 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 192 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 202 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 212 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 222 Marks
Find the expected number of boys in a family with 8 children, assuming the sex distribution to be equally probable.
Answer
Here, $\text{n}=8$
Let p be the probability of number of boys in the family.
$\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
Expected number of boys = Mean
$\Rightarrow\text{np}=4$
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Question 232 Marks
A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
none is white?
Answer
Let P be the probability of getting 1 white ball out of 7 red, 5 white and 8 black balls. so
$\text{p}=\frac{5}{20}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
Let X denote the random variable of number of selecting white ball with replacement out of 4 balls. probability of getting r white balls out of n balls is given by
$\text{P}(\text{x = r})=\text{ }^\text{n}\text{c}_\text{r}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^4\text{c}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{4-\text{r}}\dots(1)$
Probabitilty of getting none white ball
$=\text{P}(\text{x}=0)$
$=\text{ }^4\text{c}_0\big(\frac{1}{4}\big)^0\big(\frac{3}4{}\big)^{4-0}$ [Using (1)]
$=\big(\frac{3}{4}\big)^4$
$=\frac{81}{256}$
Probability of getting none white ball $=\frac{81}{256}$
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Question 242 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 252 Marks
If X follows binomial distribution with parameters n = 5, p and P(X = 2) = 9P(X = 3), then find the value of p.
Answer
We have,
X follows binomial distribution with parameters $\text{n = 5, p}$ and $\text{P(X}=2)=9\text{P(X}=3).$
So, $\text{P(X = r})=\text{ }^{5}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(5-\text{r})},$ where $\text{r}=0,1,2,3,4,5$ and $\text{q}=1-\text{p}$
As, $\text{P(X}=2)=9\text{P(X}=3)$
$\Rightarrow\text{ }^5\text{c}_2\text{p}^2\text{q}^3=9^5\text{c}_3\text{p}^3\text{q}^2$
$\Rightarrow10\text{p}^2\text{q}^3=9\times10\text{p}^3\text{q}^2$
$\Rightarrow\text{q}=9\text{p}$
$\Rightarrow1-\text{p}=9\text{p}$ [As, q = 1 - p]
$\Rightarrow10\text{p}=1$
$\therefore\text{p}=\frac{1}{10}$
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Question 262 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 272 Marks
A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
all are white?
Answer
Let P be the probability of getting 1 white ball out of 7 red, 5 white and 8 black balls. so
$\text{p}=\frac{5}{20}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
Let X denote the random variable of number of selecting white ball with replacement out of 4 balls. probability of getting r white balls out of n balls is given by
$\text{P}(\text{x = r})=\text{ }^\text{n}\text{c}_\text{r}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^4\text{c}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{4-\text{r}}\dots(1)$
Prob that all are white $=\text{P}(\text{X}=4)=\text{ }^4\text{c}_4\big(\frac{1}{4}\big)^4\big(\frac{3}{4}\big)^{4-4}=\frac{1}{256}$
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Question 282 Marks
Six coins are tossed simultaneously. Find the probability of getting.
no heads.
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 no head) = P(X = 0)
$=\frac{\text{ }^6\text{C}_0}{2^6}$
$=\big(\frac{1}{2}\big)^6$
$=\frac{1}{64}$
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Question 292 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 302 Marks
Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that
none is a spade?
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
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^5\text{c}_\text{r}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{5-\text{r}}\dots(1)$
P(none is a spade) = P(X = 0)
$=\text{ }^5\text{C}_0\big(\frac{1}{4}\big)^0\big(\frac{3}{4}\big)^5$
$=\frac{243}{1024}$
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Question 312 Marks
Find the probability distribution of the number of sixes in three tosses of a die.
Answer

$\text{X}$

$\text{P(X)}$

$0$

$\text{}^3\text{c}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{3-0}=\big(\frac{5}{6}\big)^3=\frac{125}{216}$

$1$

$\text{ }^3\text{c}_1\big(\frac{1}{6}\big)^1\big(\frac{5}{6}\big)^{3-1}=3\big(\frac{1}{6}\big)\big(\frac{1}{6}\big)^2=\frac{25}{72}$

$2$

$\text{ }^3\text{c}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{3-2}=3\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)=\frac{5}{72}$

$3$

$\text{ }^3\text{c}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^{3-3}=\big(\frac{1}{6}\big)^3=\frac{1}{216}$
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Question 322 Marks
A bag contains 2 white, 3 red and 4 blue balls. Two balls are drawn at random from the bag. If X denotes the number of white balls among the two balls drawn, describe the probability distribution of X.
Answer

$\text{X}$

$\text{P(X)}$

$0$

$\frac{7}{6}\times\frac{6}{8}=\frac{21}{36}$

$1$

$\frac{7}{9}\times\frac{2}{8}\times2=\frac{14}{36}$

$2$

$\frac{2}{9}\times\frac{1}{8}=\frac{1}{36}$
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2 Marks - Maths STD 12 Science Questions - Vidyadip