Interactive OP Malhotra ISC Class 12 Solutions Chapter 20 Theoretical Probability Distribution Ex 20(a) engage students in active learning and exploration.
S Chand Class 12 ICSE Maths Solutions Chapter 20 Theoretical Probability Distribution Ex 20(a)
Question 1.
Which of the following experiments give a discrete random variable. (You are not asked to find any probabilities).
(i) A book is chosen at random from a shelf with 5 0 books and its author noted.
(ii) A book is chosen at random from a shelf with 50 books and the number of pages noted.
(iii) A pupil is chosen at random from a particular class and the pupil’s name, is noted.
(iv) A pupil is chosen at random from a particular class and the pupil’s height is recorded to the nearest cm.
(v) The number of cars passing a given point on the road between 10: 00 and 11: 00 hours on a particular day.
Answer:
(i) Here, the set of values given to random variable is not numeric i.e. can’t be counted ∴ it is not a discrete random variable.
(ii) Here set of values which a random variable takes can be counted so it is a discrete random variable.
(iii) Here the set of values, which a random variable takes can not be counted so it is not a discrete random variable.
(iv) Here the set of values which a random variable takes can be counted. ∴ random variable is discrete.
(v) Here, the set of values which a random variable takes can be counted so it is a descrete random variable.
Question 2.
(i) Determine which of the following can be probability distribution of a random variable X:
(a)
| X | 0 | 1 | 2 |
| P(X) | 0.4 | 0.4 | 0.2 |
(b)
| X | 0 | 1 | 2 |
| P(X) | 0.5 | 0.2 | 0.2 |
(ii) Find the value of A : if a random variable X has the following probability distribution:
| X | -2 | -1 | 0 | 1 | 2 | 3 |
| P(X) | 0.1 | k | 0.2 | 2k | 0.3 | k |
Answer:
(i) (a) Here P(X = 0) + P(X = 1) + P(X = 2) = 0.4 + 0.4 + 0.2 = 1.0
Thus the sum of probabilities is equal to 1 .
Therefore, given distribution of probabilities is a probability distribution of random variable X.
(b) P(X = 0) + P(X = 1) + P(X = 2) = 0.5 + 0.2 + 0.2 = 0.9 < 1
Thus given distribution of probabilities is not a probability distribution of random variable X.
(ii) Since the sum of probabilities of probability distribution is 1 ∴ ∑ P(X) = 1
P(X = 2) + P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1
⇒ 0.1 + K + 0.2 + 2 K + 0.3 + K = 1
⇒ 4 K + 0.6 = 1
⇒ 4 K = 0.4
⇒ K = 0.1
Question 3.
Two cards are drawn in succession from a well shuffled deck of 52 cards, the first card being replaced, before the second is drawn. Let X denote the number of spades drawn. Find the probability distribution of X.
Answer:
Here, X denote the number of spade cards drawn. Then X can takes values 0,1,2.
Here, probability of getting a spade card = \(\frac{13}{52}\)
= \(\frac{1}{4}\) = p
∴ q = 1 – p = 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\)
∴ P(X = 0)
= P(no spade card ) = q
q = \(\frac{3}{4}\) × \(\frac{3}{4}\)
= \(\frac{9}{16}\)
P(X = 1) = P (one spade card in two draws)
= p q + q p = \(\frac{1}{3}\) × \(\frac{3}{4}\) + \(\frac{3}{4}\) × \(\frac{1}{4}\)
= \(\frac{6}{16}\)
P(X = 2) = p p = \(\frac{1}{4}\) × \(\frac{1}{4}\)
= \(\frac{1}{16}\)
Thus the probabilities distribution of X is given below:
| X | 0 | 1 | 2 |
| P(X) | \(\frac{9}{16}\) | \(\frac{6}{16}\) | \(\frac{1}{16}\) |
Question 4.
Shew graphically the probability function of the discrete random variable X, where X is the number of heads appearing when an unbiased coin is tossed twice in succession.
Answer:
Since X is the no, of heads appearing when an unbiased coin is tossed twice
∴ Sample space = {H.H, HT, TH, TT}
Thus X can takes values 0, 1, 2
P(X = 0) = P(T T) = \(\frac{1}{4}\) ;
P(X = 1) = P(H T, T H) = \(\frac{2}{4}\) = \(\frac{1}{2}\) ;
P(X = 2) = P(H H) = \(\frac{1}{4}\)
Find the probability distribution of X in questions 5-7. Sketch the graphs.
Question 5.
A box contains 3 red marbles and 5 green marbles. Two marbles are taken at random without replacement and X is the number of green marbles obtained.
Answer:
Given X denote the no. of green marbles obtained so X can take values 0,1,2.
P(X = 0) = prob. of getting no green marble and 2 red marbles = \(\frac{{ }^3 \mathrm{C}_2}{{ }^8 \mathrm{C}_2}\)
= \(\frac{3}{\frac{8 \times 7}{2}}\)
= \(\frac{3}{28}\)
P(X = 1) = prob. of getting 1 green and I red marble
= \(\frac{{ }^3 \mathrm{C}_1 \times{ }^5 \mathrm{C}_1}{{ }^8 \mathrm{C}_2}\)
= \(\frac{3 \times 5}{\frac{8 \times 7}{2}}\)
= \(\frac{15}{28}\)
P(X = 2) = prob. of getting 2 green marables = \(\frac{{ }^5 \mathrm{C}_2}{{ }^8 \mathrm{C}_2}\)
= \(\frac{\frac{5 \times 4}{2}}{\frac{8 \times 7}{2}}\)
= \(\frac{20}{56}\) = \(\frac{5}{14}\)
= \(\frac{10}{28}\)
Thus the probability distribution of X is given below :
| X | 0 | 1 | 2 |
| P(X) | \(\frac{3}{28}\) | \(\frac{15}{28}\) | \(\frac{10}{28}\) |
Question 6.
A fair coin has, the number ‘ 1 ‘ on one face and the number ‘ 2 ‘ on the other. The coin is thrown with a fair die and X is the sum of the scores.
Answer:
Since a fair coin has number 1 on one face and 2 on other face.
When the coin and dice is thrown and X be the random variable denotes the sum of scores.
So X = 2, 3, 4, 5, 6, 7, 8.
Here total no. of outcomes = 12
{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\}
P(X = 2) = P(1,1) = \(\frac{1}{12}\) ;
P(X = 3) = P(1,2),(2,1) = \(\frac{2}{12}\)
P(X = 4) = P(1,3),(2,2) = \(\frac{2}{12}\) ;
P(X = 5) = P(1,4),(2,3) = \(\frac{2}{12}\)
P(X = 6) = P(1,5),(2,4) = \(\frac{2}{12}\) ;
P(X = 7) = P(1,6),(2,5) = \(\frac{2}{12}\)
P(X = 8) = P(2,6) = \(\frac{1}{12}\)
Thus the probability distribution of X is given below:
| X | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X) | \(\frac{1}{12}\) | \(\frac{2}{12}\) | \(\frac{2}{12}\) | \(\frac{2}{12}\) | \(\frac{2}{12}\) | \(\frac{2}{12}\) | \(\frac{1}{12}\) |
Question 7.
An urn contains 4 white and 3 red balls. Find the probability distribution of the number of red balls in a random draw of three balis.
Answer:
Let X denote the random variable represents the no. of red balls in a random draw of three balls. ∴ X can takes values 0, 1, 2, 3.
Since the urn contains 4 white and 3 red balls
P(X = 0) = prob. of drawing 3 white balls
= \(\frac{{ }^4 \mathrm{C}_3}{{ }^7 \mathrm{C}_3}\)
= \(\frac{4}{\frac{7 \times 6 \times 5}{6}}\)
= \(\frac{4}{35}\)
P(X = 1) = prob. of drawing 1 red ball and 2 white balls = \(\frac{{ }^4 C_2 \times{ }^3 C_1}{{ }^7 C_3}\)
= \(\frac{\frac{4 \times 3}{2} \times 3}{\frac{7 \times 6 \times 5}{6}}\)
= \(\frac{18}{35}\)
P(X = 2) = prob. of drawing 2 red balls and 1 white ball
= \(\frac{{ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2}{{ }^7 \mathrm{C}_3}\)
= \(\frac{4 \times 3}{\frac{7 \times 6 \times 5}{6}}\)
= \(\frac{12}{35}\)
P(X = 3) = prob. of drawing 3 red balls
= \(\frac{{ }^3 \mathrm{C}_3}{{ }^7 \mathrm{C}_3}\)
= \(\frac{1}{35}\)
Thus the probability distribution of X is given below:
| X | 0 | 1 | 2 | 3 |
| P(X) | \(\frac{4}{35}\) | \(\frac{18}{35}\) | \(\frac{12}{35}\) | \(\frac{1}{35}\) |
Question 8.
Two cards are drawn successively without replacement from a well shuffled pack of 52 cards. Find the probability distribution of the jmmber of aces.
Answer:
Let X be the random variable denotes the number of aces in two draws without replacement.
Thus X can take values 0,1,2.
∴ P(X = 0) = Prob. of getting no ace = \(\frac{48}{52}\) × \(\frac{47}{51}\)
= \(\frac{12}{13}\) × \(\frac{47}{51}\)
= \(\frac{4 \times 47}{13 \times 17}\)
= \(\frac{188}{221}\)
P(X = 1) = prob. of getting one ace and one other card
= \(\frac{2 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_1}\) × \(\frac{{ }^{48} \mathrm{C}_1}{{ }^{51} \mathrm{C}_1}\)
= \(\frac{2 \times 4 \times 48}{52 \times 51}\)
= \(\frac{2 \times 1 \times 16}{13 \times 17}\)
= \(\frac{32}{321}\)
P(X = 2) = prob. of getting two ace cards in two draws
= \(\frac{4}{52}\) × \(\frac{3}{51}\)
= \(\frac{1}{221}\)
| X | 0 | 1 | 2 |
| P(X) | \(\frac{188}{221}\) | \(\frac{32}{321}\) | \(\frac{1}{221}\) |
Thus probability distribution of X is given by
Question 9.
From a list containing 25 items, 5 of which are defective, 4 are chosen at random. Let X be the number of defectives found. Obtain the probability distribution of X, if the items are chosen without replacement.
Answer:
Total no. of items = 25
no. of defective items = 5
∴ no. of non-defective items = 25 – 5 = 20
Let X be the random variable denotes the no. of defective item found.
∴ X can take values 0,1,2,3,4.
P(X = 0) = prob. of getting no defective items i.e. getting 4 non-defective items
= \(\frac{{ }^{20} \mathrm{C}_4}{{ }^{25} \mathrm{C}_4}\)
= \(\frac{20 \times 19 \times 18 \times 17}{25 \times 24 \times 23 \times 22}\)
= \(\frac{969}{2530}\)
P(X= 1) = prob. of getting one defective and 3 non-defective items
= \(\frac{{ }^5 \mathrm{C}_1 \times{ }^{20} \mathrm{C}_3}{{ }^{25} \mathrm{C}_3}\)
= \(\frac{5 \times \frac{20 \times 19 \times 18}{6}}{\frac{25 \times 24 \times 23 \times 22}{24}}\)
= \(\frac{5 \times 20 \times 19 \times 3}{25 \times 23 \times 22}\)
= \(\frac{114}{253}\)
P(X = 2) = prob. of getting = defective and 2 non-defective item
= \(\frac{{ }^5 \mathrm{C}_2 \times{ }^{20} \mathrm{C}_2}{{ }^{25} \mathrm{C}_4}\)
= \(\frac{\frac{5 \times 4}{2} \times \frac{20 \times 19}{2}}{\frac{25 \times 25 \times 23 \times 22}{24}}\)
= \(\frac{5 \times 20 \times 19}{25 \times 23 \times 22}\)
= \(\frac{38}{253}\)
P(X = 3) = prob. of getting 3 defective and 1 non-defective item
= \(\frac{{ }^5 \mathrm{C}_3 \times{ }^{20} \mathrm{C}_1}{{ }^{25} \mathrm{C}_4}\)
= \(\frac{\frac{5 \times 4}{2} \times 20}{25 \times 23 \times 22}\)
= \(\frac{4}{253}\)
P(X = 4) = prob. of getting 4 defective items
= \(\frac{{ }^5 \mathrm{C}_4}{{ }^{25} \mathrm{C}_4}\)
= \(\frac{5}{25 \times 23 \times 22}\)
= \(\frac{1}{2530}\)
Thus the probability distribution of random variable X is given by
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | \(\frac{969}{2530}\) | \(\frac{114}{253}\) | \(\frac{38}{253}\) | \(\frac{4}{253}\) | \(\frac{1}{2530}\) |
Question 10.
A random variable A ” has the following probability distribution.
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X) | a | 3a | 5a | 7a | 9a | 11a | 13a | 15a | 17a |
Determine:
(i) The value of a.
(ii) P(X < 3) (iii) P(X > 3)
(iv) P(0 < X < 5)
Answer:
(i) Since ∑P(X) = 1
⇒ P(X= 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 1
⇒ a + 3 a + 5 a + 7 a + 9 a + 11 a + 13 a + 15 a + 17 a = 1
⇒ 81 a = 1
⇒ a = \(\frac{1}{81}\)
(ii) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= a + 3 a + 5 a = 9 a
= \(\frac{9}{81}\)
= \(\frac{1}{9}\)
(iii) P(X < 3) = 1 – P(X < 3)
= 1 – \(\frac{1}{9}\)
= \(\frac{8}{9}\)
(iv) P(0 < X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 3 a + 5 a + 7 a + 9 a = 24 a
= \(\frac{24}{81}\)
= \(\frac{8}{27}\)
Question 11.
A random variable X has the following probability distribution:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | a | 4a | 3a | 7a | 8a | 10a | 6a | 9a |
(i) Find the value of a;
(ii) Find P(X < 3), P(X < 4), P(0 < X < 5);
(iii) Give the smallest value of m for which P(X > m) < 0.6.
Answer:
(i) For probability distribution, ∑ P(X) = 1
i.e. P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 1
⇒ a+4 a+3 a+7 a+8 a+10 a+6 a+9 a = 1
⇒ 48 a = 1
⇒ a = \(\frac{1}{48}\)
(ii) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = a + 4 a + 3 a = 8 a = \(\frac{8}{48}\) = \(\frac{1}{6}\) = 0.167 P(X > 4) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)
= 8 a+10 a+6 a+9 a
= 33 a
= \(\frac{33}{48}\)
= \(\frac{11}{16}\)
= 0.6875
and P(0 < X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 4 a + 3 a + 7 a + 8 a = 22 a
= \(\frac{22}{48}\)
= \(\frac{11}{24}\)
= 0.458
(iii) Given P(X < m) > 0.6
The probability distribution of X is given by
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | \(\frac{1}{48}\) | \(\frac{4}{48}\) = 0.0.83 | \(\frac{3}{48}\) = \(\frac{1}{16}\) | \(\frac{7}{48}\) = 0.145 | 0.167 | 0.2083 | 0.125 | 0.1875 |
Clearly P(X < 4)
= a + 4 a + 3 a + 7 a + 8 a = 23 a
= \(\frac{23}{48}\)
= 0.479
P(X < 5) = a + 4 a + 3 a + 7 a + 8 a + 10 a
= 33 a
= \(\frac{33}{48}\)
= 0.6875
Thus the smallest value of m for which P(X < m) > 0.6 be 5 .
Since P(X < 5) = 0.6875 > 0.6
But P(X < 4) = 0.479 < 0.6