Probability Distribution of a Random Variable:
In statistics, we had sufficient discussions about frequency distribution, which were based on observations. In a frequency distribution, the frequencies for different values of the variable under consideration are based on actual observation. Thus, if an unbiased coin is tossed 20 times, we may get head 13 times, though theoretically, we shall expect head 10 times. But 13 is the observed frequency here.
In this section, we shall discuss probability distribution based on theoretical considerations. A symbol that can assume any of the prescribed set of values is known as a variable. We consider S to be the Sample Space of some given random experiment. The outcomes, i.e., the points of the sample space are not always numbers. But we may assign a real number to each sample point according to some definite rule, which gives us a function defined on the sample space S. This real-valued function defined on the sample space of an experiment is known as a random variable. The values of a random variable are real numbers connected with the outcomes of an experiment.
In the random experiment of tossing 3 coins, the sample space is given by-
S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }
If x be the random variable denoting the ‘number of heads’ in this case, then we assign a number to each sample point as follows:
| Sample Points | HHH | HHT | HTH | THH | HTT | THT | TTH | TTT |
| x | 3 | 2 | 2 | 2 | 1 | 1 | 1 | 0 |
One can define many other random variables on the same sample space. For example, if x denotes the random variable defined as the ‘square of the number of tails’ in the above experiment, then we have
| Sample Points | HHH | HHT | HTH | THH | HTT | THT | TTH | TTT |
| x | 0 | 1 | 1 | 1 | 4 | 4 | 4 | 9 |
In the first example, the random variable x can assume only four discrete values 0, 1, 2, and 3 and in the second example, x can assume the values 0, 1, 4, and 9. The random variables defined in these two cases are called discrete random variables. We can define a discrete random variable as the one which can assume only finite number of values. If, however, the random variable assumes any value between certain limits, then it is known as a continuous random variable.
| Example 1- Find the probability distribution of the number of tails when three coins are tossed. Solution- Let ‘X’ denoted the number of tails when 3 coins are tossed. So, X can take values 0, 1, 2, 3. Let ‘S’ be the sample space of tossing of 3 coins. S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT } n(S) = 8 P(x = 0) = P(no tails) = 1/8 P(x = 1) = P(one tail) = 3/8 P(x = 2) = P(two tail) = 3/8 P(x = 3) = P(three tail) = 1/8 |
The required probability distribution is-
| X | 0 | 1 | 2 | 3 |
| P(X) | 1/8 | 3/8 | 3/8 | 1/8 |
| Example 2- Two cards are drawn successively, with replacement, from a well-shuffled pack of 52 cards. Find the probability distribution of the number of kings. Solution- Let X be the number of kings drawn from 52 cards one by one with replacement. So, X can take values 0, 1, 2. Now, P(X = 0) = P(no king) ⇒ P(X = 0) = P(Ā1 ∩ Ā2) ⇒ P(X = 0) = P(Ā1) . P(Ā2) = (48/52) (48/52) = 144/169 P(X = 1) = P(getting 1 king) ⇒ P(X = 1) = P(king in the first card and not a king in the second card) + P(not a king in the first card and king in the second card) ⇒ P(X = 1) = P[ (A1 ∩ Ā2) + (Ā1 ∩ A2) ] ⇒ P(X = 1) = P(A1 ∩ Ā2) + P(Ā1 ∩ A2) ⇒ P(X = 1) = P(A1) P(Ā2) + P(Ā1) P(A2) ⇒ P(X = 1) = (4/52) (48/52) + (48/52) (4/52) ⇒ P(X = 1) = (1/13) (12/13) + (12/13) (1/13) = 24/169 P(X = 2) = P(Both are king) ⇒ P(X = 2) = P(A1 ∩ A2) ⇒ P(X = 2) = P(A1) P(A2) ⇒ P(X = 2) = (4/52) (4/52) = 1/169 |
The required probability distribution is-
| X | 0 | 1 | 2 |
| P(X) | 144/169 | 24/169 | 1/169 |
| Example 3- Two cards are drawn from a well-shuffled deck of 52 cards. Find the probability distribution of the number of kings, if the cards are drawn at random. Solution- Let X be the number of kings drawn from 52 cards at random. So, X can take values 0, 1, 2. P(X = 0) = P(no king) = 48C2/52C2 = 188/221 P(X = 1) = P(one king) ⇒ P(X = 1) = P(one king and other is non-king) ⇒ P(X = 1) = (4C1 x 48C1)/52C2 = 32/221 P(X = 2) = P(Both are king) = 4C2/52C2 = 1/221 |
The required probability distribution is-
| X | 0 | 1 | 2 |
| P(X) | 188/221 | 32/221 | 1/221 |
| Example 4- Find the probability distribution for the number of girls in a family with three children, assuming equal probabilities for a child being a boy or a girl. Solution- Let ‘X’ be the number of girls in a family of three children. So X can take values 0, 1, 2, 3. Let S be the sample space of 3 children in a family. S = { GGG, GGB, GBG, BGG, GBB, BGB, BBG, BBB } n(S) = 8 P(X = 0) = P(no girls) = 1/8 P(X = 1) = P(1 girl) = 3/8 P(X = 2) = P(2 girls) = 3/8 P(X = 3) = P(all are girls) = 1/8 |
The required probability distribution is-
| X | 0 | 1 | 2 | 3 |
| P(X) | 1/8 | 3/8 | 3/8 | 1/8 |
| Example 5- Two bad eggs are accidentally mixed with 10 good ones. Find the probability distribution of the number of bad eggs in 3 draws at random, one by one, without replacement, from this lot. Solution- Let X denote the number of bad eggs in 3 draws at random, one by one. Then, X = 0, 1, 2 (as the number of bad eggs = 2). Also, the number of good eggs = 10, and the total number of eggs = 12. ∴ P(X = 0) = P(no bad egg) = (10/12) (9/11) (8/10) = 12/22 P(X = 1) = P(1 bad egg and 2 good ones) ⇒ P(X = 1) = P(BGG) + P(GBG) + P(GGB) ⇒ P(X = 1) = (2/12) (10/11) (9/10) + (10/12) (2/11) (9/10) + (10/12) (9/11) (2/10) = 9/22 P(X = 2) = P(2 bad eggs and 1 good) ⇒ P(X = 2) = P(BBG) + P(BGB) + P(GBB) ⇒ P(X = 2) = (2/12) (1/11) (10/10) + (2/12) (10/11) (1/10) + (10/12) (2/11) (1/10) = 1/22 |
The required probability distribution is-
| X | 0 | 1 | 2 |
| P(X) | 12/22 | 9/22 | 1/22 |
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