Probability Distribution PDF

## PROBABILITY **15.60** **Note:** If *x* is one of the possible values of a random variable *X*, then the probability that *X* takes value *x* is always non-zero i.e., *P(X=x)≠0*. ### SOLVED EXAMPLES **EXAMPLE 1.** A person plays a game of tossing a coin thrice. For each head, he gains ₹5, and for...

## PROBABILITY **15.60** **Note:** If *x* is one of the possible values of a random variable *X*, then the probability that *X* takes value *x* is always non-zero i.e., *P(X=x)≠0*. ### SOLVED EXAMPLES **EXAMPLE 1.** A person plays a game of tossing a coin thrice. For each head, he gains ₹5, and for each tail, he loses ₹2. Let *X* denote the amount gained or lost by the person. Show that *X* is a random variable and exhibit it as a function on the sample space of the experiment. **Solution.** Let *S* be the sample space of the experiment i.e., when a coin is tossed thrice, *S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}*. Now, it is given that the person gains ₹5 on getting each head and loses ₹2 on getting each tail. Here *X* is the amount gained or lost by the person and its values depend on the outcomes of the random experiment which are as follows: * *X(HHH) = ₹(3×5) = ₹ 15* * *X(HHT) = X(HTH) = X(THH) = ₹(2×5−1×2) = ₹ 8* * *X(HTT) = X(THT) = X(TTH) = ₹(1×5−2×2) = ₹ 1* * *X(TTT) = −₹(3×2) = −₹6* Thus, *X* is a random variable. Also, for each element of the sample space, *X* assumes a unique value. Hence *X* is a function on the sample space of the experiment having range {-6,1,8,15}. **EXAMPLE 2.** Find the probability distribution of the number of heads in three tosses of a coin. **Solution.** When a coin is tossed three times, the possible outcomes (denoting head by *H* and tail by *T*) are TTT, TTH, THT, HTT, THH, HTH, HHT, and HHH. Let *X* denote the random variable ‘number of heads when a coin is tossed thrice’. Clearly, the possible values of *X* are 0, 1, 2, 3. Now, * *P(X=0) = Probability of getting no head = P(TTT) = 1/8* * *P(X=1) = Probability of getting one head = P(TTH, THT, HTT) = 3/8* * *P(X=2) = Probability of getting 2 heads = P(THH, HTH, HHT) = 3/8* * *P(X=3) = Probability of getting three heads = P(HHH) = 1/8* Thus, the required probability distribution is: * *X:* 0 1 2 3 * *P(X):* 1/8 3/8 3/8 1/8 **EXAMPLE 3.** Find the probability distribution of a number of successes in two tosses of a die, where a success is defined as a number greater than 4. Sketch its graph also. **Solution.** Let *S* denote the number greater than 4 and failure *F* (say) is defined as a 'number less than 4'. Clearly, *X* can take the values 0, 1, 2. * *P(S) = P(5 or 6) = 2/6 = 1/3* * *P(F) = P(1 or 2 or 3 or 4) = 4/6 = 2/3* Now, * *P(X=0) = P(no success) = P(FF) = P(F).P(F) = 2/3 × 2/3 = 4/9* * *P(X=1) = P(one success) = P(SF or FS) = P(SF) + P(FS)* * * = P(S)P(F) + P(F)P(S) = 1/3 × 2/3 + 2/3 × 1/3 = 1/9 + 1/9 = 2/9* * *P(X=2) = P(both successes) = P(SS) = P(S).P(S) = 1/3 × 1/3 = 1/9* Hence, the required probability distribution is: * *X:* 0 1 2 * *P(X):* 4/9 2/9 1/9 **EXAMPLE 4.** A bag contains 3 white and 4 red balls. Three balls are drawn one by one with replacement. Find the probability distribution of the number of red balls. **Solution.** Let *X* denote the number of red balls drawn. Now, 3 balls are drawn one by one with replacement from the bag. Clearly, *X* can take the values 0, 1, 2, 3. Let *S* denote the event of 'drawing a red ball' and *F* denote the event of 'not drawing a red ball'. Then, * *P(S) = 4/7* and *P(F) = 3/7* Now, * *P(X=0) = P(FFF) = P(F)P(F)P(F) = 3/7 × 3/7 × 3/7 = 27/343* * *P(X=1) = P(SFF or FSF or FFS) = P(SFF)+P(FSF)+P(FFS)* * * = P(S)P(F)P(F) + P(F)P(S)P(F) + P(F)P(F)P(S)* * * = 4/7 × 3/7 × 3/7 + 3/7 × 4/7 × 3/7 + 3/7 × 3/7 × 4/1 = 36 / 343 = 108/343* * *P(X=2) = P(SSF or SFS or FSS) = P(SSF) + P(SFS) + P(FSS)* * * = P(S)P(S)P(F) + P(S)P(F)P(S) + P(F)P(S)P(S)* * * = 4/7 × 4/7 × 3/7 + 4/7 × 3/7 × 4/7 + 3/7 × 4/7 × 4/7 = 48 / 343 = 144/343* * *P(X=3) = P(SSS) = P(S)P(S)P(S)* * * = 4/7 × 4/7 × 4/7 = 64/343* Hence, the required probability distribution is: * *X:* 0 1 2 3 * *P(X):* 27/343 108/343 144/343 64/343 **EXAMPLE 5.** An urn contains 4 white and 6 red balls. Four balls are drawn at random from the urn . Find the probability distribution of the number of white balls. [CBSE 2012C] **Solution.** Let *X* denote the number of white balls drawn. Since there are 4 white balls in the urn, therefore *X* can take the values 0, 1, 2, 3, and 4. Now, * *P(X=0) = Probability that all 4 balls drawn are red = 6C4 / 10C4 = 1/14* * *P(X=1) = Probability of drawing one while ball = 4C1 × 6C3 / 10C4 = 8/21* * *P(X=2) = Probability of drawing two white balls = 4C2 × 6C2 / 10C4 = 6/14* * *P(X=3) = Probability of drawing three white balls = 4C3 × 6C1 / 10C4 = 4/35* * *P(X=4) = Probality of drawing 4 white balls = 4C4 × 6C0 / 10C4 = 1 / 210* Hence, the required probability distribution is: * *X:* 0 1 2 3 4 * *P(X):* 1/14 8/21 6/14 4/35 1/210 **EXAMPLE 6.** Two bad eggs are mixed accidentally with 10 good ones. Find the probability distribution of the number of bad eggs in 3 draws at random, without replacement, from this lot. [CBSE 2015C] **Solution.** Let *X* denote the number of bad eggs obtained in 3 draws from a group of 10 good and 2 bad eggs. Clearly, *X* can take the values 0, 1, or 2. Total number of eggs=12. * *P(X=0) = Probability of drawing no bad eggs = 10C3 / 12C3 = 12/22* * *P(X=1) = Probability of drawing one bad egg and two good eggs = 2C1 × 10C2 / 12C3 = 9/22* * *P(X=2) = Probability of drawing two bad eggs and one good egg = 2C2 × 10C1 / 12C3 = 1/22* Hence, the required probability distribution is: * *X:* 0 1 2 * *P(X):* 12/22 9/22 1/22 **EXAMPLE 7.** Three numbers are selected at random (without replacement) from first six positive integers. Let *X* denote the largest of the three numbers obtained. Find the probability distribution of *X*. **Solution.** Here *X* denotes 'largest of the three numbers obtained', when three numbers are selected, without replacement, from numbers 1, 2, 3, 4, 5, 6. Clearly, *X* is a random variable which can take the values 3, 4, 5, 6. Then, * *P(X=3) = Probality that the largest of three numbers is 3 = P(selecting 3 and any two out of 1, 2)* * * = 1C1 × 2C2 / 6C3 = 1/20* * *P(X=4) = Probability that the largest of the three numbers is 4 = P(selecting 4 and any two out of 1, 2, 3)* * * = 1C1 × 3C2 / 6C3 = 3/20* * *P(X=5) = Probability that the largest of the three numbers is 5 = P(selecting 5 and any two out of 1, 2, 3, 4)* * * = 1C1 × 4C2 / 6C3 = 6/20* * *P(X=6) = Probability that the largest of the three numbers is 6 = P(selecting 6 and any two out of 1, 2, 3, 4, 5)* * * = 1C1 × 5C2 / 6C3 = 10/20* Hence, the required probability distribution is: * *X:* 3 4 5 6 * *P(X):* 1/20 3/20 6/20 10/20 **EXAMPLE 8.** A random variable *X* has the following probability distribution values of *X*: * *X:* 0 1 2 3 4 5 6 7 * *P(X):* 0 *k* 2*k* 2*k* 3*k* *k*² 2*k*² 7*k*² + *k* Find (i) *k* (ii) *P(X<3)* (iii) *P(X>6)* (iv) *P(X<6)* (v) *P(0<X<5)* **Solution.** (i) We know that the sum of all the probabilities in a probability distribution is always 1. * *P(X=0) + P(X=1) +...+ P(X=7) = 1* * *0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1* * *10k²+9k−1=0* * *(10k−1)(k+1)=0* * *10k−1=0 => k=1/10* *k+1≠0 => k≠−1* (ii) *P(X<3) = P(X=0) + P(X=1) + P(X=2)* * * = 0 + k + 2k = 3k = 3×1/10 = 3/10* (iii) *P(X>6) = P(X=7) = 7k²+k* * * = 7(1/10)² + 1/10 = 7/100 + 1/10 = 17/100* (iv) *P(X<6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)* * = 0 + k + 2k + 2k + 3k + k² = k² + 8k* * = (1/10)² + 8×1/10 = 1/100 + 8/10 = 81/100* (v) *P(0<X<5) = P(X=1) + P(X=2) + P(X=3) + P(X=4)* * = k + 2k + 2k + 3k = 8k = 8×1/10 = 4/5* **EXAMPLE 9.** Let *X* be a random variable which assumes values *x₁*, *x₂*, *x₃*, *x₄* such that 2*P(X=x₁)*= 3*P(X=x₂)* = *P(X=x₃)* = 5*P(X=x₄)*. Find the probability distribution of *X*. **Solution.** Here, *X* is a random variable which assumes the values *x₁*, *x₂*, *x₃*, and *x₄* such that 2*P(X=x₁)*= 3*P(X=x₂)* = *P(X=x₃)* = 5*P(X=x₄)*. Let *P(X=x₃)* = *k*, then * *P(X=x₁) = k/2*, *P(X=x₂)= k/3* and *P(X=x₄)= k/5* [: Sum of probabilities = 1] * *k/2 + k/3 + k + k/5= 1* * *15k + 10k + 30k + 6k / 30 = 1* * *61k/30 = 1 => k=30/61* Hence, the required probability distribution is * *X:* *x₁* *x₂* *x₃* *x₄* * *P(X):* 15/61 10/61 30/61 6/61 **EXAMPLE 10.** Let *X* denote the number of hours you study during a randomly selected school day. The probability that *X* can take the values *x* has the following form, where *k* is some unknown constant. * *P(X=x)*= * *0.1 if x=0* * *kx if x=1 or 2* * *k(5−x) if x=3 or 4* * *0 otherwise* **(a)** Find the value of *k*. **(b)** What is the probability that you study (i) atleast two hours (ii) atmost two hours? **Solution.** The probability distribution for the random variable *X* is * *X:* 0 1 2 3 4 * *P(X):* 0.1 *k* 2*k* 2*k* *k* **(a)** We know that the sum of the probabilities in a probability distribution is always 1. * *P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1* * *0-1 + k + 2k + 2k + k = 1* * *0-1 + 6k = 1 => 6k = 0.9 => k=0.15* **(b)** (i) *P(studying atleast two hours) = P(X≥2)* * * = P(X=2) + P(X=3) + P(X=4)* * * = 2k + 2k + k = 5k = 5×0.15 = 0.75* (ii) *P(studying atmost two hours) = P(X≤2)* * * = P(X=0) + P(X=1) + P(X=2)* * * = 0.1 + k + 2k = 0.1 + 3k* * * = 0.1 + 3×0.15 = 0.1+0.45 = 0.55* ### **EXERCISE 15.7** 1. Find the probability distribution of the random variable ‘number of tails’ when two coins are tossed. [NCERT] 2. Find the probability distribution of the number of tails in four tosses of a coin. 3. A coin is biased so that the head is 3 times as likely to occur as a tail. If the coin is tossed twice, find the probability distribution for the number of tails. [AICBSE 2020; CBSE Sample Paper 2010] 4. Find the probability distribution of the number of times a total of 9 appears in two throws of two dice. 5. Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of queens. [CBSE 2011C, 08, Sample Paper 2000] 6. Two cards are drawn simultaneously from a pack of 52 cards without replacement. Find the probability distribution of the number of kings. [CBSE Sample Paper 2004] 7. Three cards are drawn successively with replacement from a well shuffled deck of 52 cards. A random variable *X* denotes the number of spades on three cards. Determine the probability distribution of *X*. [CBSE 2009 C] 8. An urn contains 4 white and 3 red balls. Find the probability distribution of the number of red balls in three draws, with replacement, from the urn. 9. 3 bad articles are mixed with 7 good ones. Find the probability distribution of bad articles if 2 are drawn at random without replacement from this lot. 10. Three balls are drawn without replacement from a bag containing 5 white and 4 red balls. Find the probability distribution of the number of red balls drawn. [CBSE Sample Paper 2021] 11. A box contains 13 bulbs out of which 5 bulbs are defective. 3 bulbs are drawn one by one from the box without replacement. Find the probability distribution of the number of defective bulbs drawn. 12. A pair of unbiased dice is thrown. If the random variable *X* is the sum of two numbers obtained on two dice, find the probability distribution of *X*. 13. Find the probability distribution of the number of doublets in three throws of a pair of dice. [CBSE 2010C; CBSE 2011C, 05C] 14. From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random. Let the random variable *X* denote the number of defective items in the sample. If the sample is drawn without replacement, find (i) the probability distribution of *X* (ii) *P(X≤1)* (iii) *P(X<1)* (iv) *P(0<X<2)*. 15. The probability distribution function of a random variable *X* is given by * *X:* 0 1 2 * *P(X)*: 3*k*³ 4*k* − 10*k*² 5*k* − 1 where *k* > 0. Find (i) *k* (ii) *P(X<2)* (iii) *P(1<X≤2)* 16. A random variable *X* has the following probability distribution: * *X:* 0 1 2 3 4 5 6 7 8 * *P(X)*: *k* 3*k* 5*k* 7*k* 8*k* 11*k* 0 *k* 3*k* Find (i) *P(X<3)* (ii) *P(2≤X≤5)* 17. The random variable *X* has a probability distribution *P(X)* of the following form, where ‘*k*’ is some real number. * *P(X)* = * *k*, if *x*=0 * *2k*, if *x*=1 * *3k*, if *x*=2 * *0* otherwise (i) Determine the value of *k*. (ii) Find *P(X<2)* (iii) Find *P(X>2)* [CBSE Sample Paper 2024] 18. Let *X* denote the number of colleges where you will apply after your results and *P(X=x)* denotes your probability of getting admission in *x* number of colleges. It is given that * *P(X=x)* = * *kx*, if *x*=0 or 1 * *2kx*, if *x*=2 * *k(5−x)*, if *x*=3 or 4, where *k* is a positive constant (i) Find the value of *k*. (ii) What is the probability that you will get admission in exactly two colleges? (iii) Find the mean and variance of the probability distribution. [CBSE 2016] ## **ANSWERS** 1. *X:* 0 1 2 * *P(X):* 1/4 1/2 1/4 2. *X:* 0 1 2 3 4 * *P(X):* 1/16 1/4 3/8 1/4 1/16 3. *X:* 0 1 2 * *P(X):* 1/16 6/16 9/16 4. *X:* 0 1 * *P(X):* 64/81 16/81 1/81 5. *X:* 0 1 2 * *P(X):* 144/169 24/169 1/169 6. *X:* 0 1 2 * *P(X):* 188/221 32/221 1/221 7. *X:* 0 1 2 3 * *P(X):* 27/64 27/64 9/64 1/64 8. *X:* 0 1 2 3 * *P(X):* 64/343 144/343 108/343 27/343 9. *X:* 0 1 2 3 * *P(X):* 7/24 21/40 7/40 1/120 10. *X:* 0 1 2 3 * *P(X):* 42/286 140/286 80/286 10/286 11. *X:* 1 2 3 4 5 6 7 8 9 10 11 12 * *P(X):* 36/36 36/36 36/36 36/36 36/36 36/36 36/36 36/36 36/36 36/36 36/36 36/36 12. *X:* 0 1 2 3 * *P(X):* 125/216 75/216 15/216 1/216 13. *X:* 0 1 2 3 * *P(X):* 1/6 1/2 1/3 1/30 14. *X:* 0 1 2 * *P(X):* 1/6 2/10 3/30 15. (i) *k* = 1/3 (ii) *P(X<2)* = 1 (iii) *P(1<X≤2)* = 2/3 16. (i) *k* = 1/34 (ii) *P(X<3)* = 3/17 (ii) *P(2≤X≤5)* = 5/17 17. (i) *k* = 1/8 (ii) *P(X<2)* = 3/8 (iii) *P(X>2)* = 1/4 18. (i) *k* = 1/8 (ii) *P(X=2)* = 1/4 (iii) Mean = 2; Variance = 1 1. (i) Mean = 2; Var. = 1 (ii) Mean = −0.2; Var. = 1.56 (iii) Mean = 3.8; Var. = 2.36 2. Mean = 7/2; Var. = 2.92, S.D. = 1.71 3. Mean = 2; Var. = 1 4. Mean = 1; Var. = 1/2 5. (i) *X:* 0 1 2 * *P(X):* 25/36 10/36 1/36 (ii) *X:* 0 1 2 3 * *P(X):* 125/216 75/216 15/216 1/216 Mean; Var. = 5/3, 5/18 Mean = 2; Var. = 5/12 6. 2/3 7. (i) *X:* 0 1 2 3 * *P(X):* 9/16 6/16 1/16 1/16; Mean = 1/2; Var. = 3/8 (ii) *X:* 0 1 2 3 * *P(X):* 1728/2197 432/2197 36/2197 1/2197; Mean = 13/13 8. (i) Mean = 3/7; Var. = 400/2873 (ii) Mean = 1; Var. = 0.49 9. Mean = 2.1; Var. = 0.79 10. *X:* 0 1 2 3 * *P(X):* 1/27 6/27 12/27 8/27; Mean=2 11. *X:* 0 1 2 4 * *P(X):* 1/4 1/2 1/4 1/4; Mean=1 12. 1 40 13. 3/4 14. *X:* 0 1 2 * *P(X):* 16/25 8/25 1/25; Mean=1/5 16. Mean = 4; Variance = 1 15. *X:* 0 1 2 3 4 5 6 7 8 9 10 * *P(X):* 343/1000 441/1000 189/1000 27/1000 1/1000 17. *X:* 1 2 3 4 5 6 * *P(X):* 5/21 4/21 3/21 2/21 2/21 1/21; Mean = 3 18. (i) *k* = 1/8 (ii) (iii) Mean = 5/8; Var. = 64/47