Chapter 13 Probability Ex 13.4 Solutions
Question - 1 : - State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
Answer - 1 : - (i)
Here we have tablewith values for X and P(X).
As we know the sum ofall the probabilities in a probability distribution of a random variable mustbe one.
Hence the sum ofprobabilities of given table = 0.4 + 0.4 + 0.2
= 1
Hence, the given tableis the probability distributions of a random variable.
(ii)
Here we have tablewith values for X and P(X).
As we see from thetable that P(X) = -0.1 for X = 3.
It is known thatprobability of any observation must always be positive that it can’t benegative.
Hence, the given tableis not the probability distributions of a random variable.
(iii)
Here we have tablewith values for X and P(X).
As we know the sum ofall the probabilities in a probability distribution of a random variable mustbe one.
Hence the sum ofprobabilities of given table = 0.6 + 0.1 + 0.2
= 0.9 ≠ 1
Hence, the given tableis not the probability distributions of a random variable.
(iv)
Here we have tablewith values for X and P(X).
As we know the sum ofall the probabilities in a probability distribution of a random variable mustbe one.
Hence the sum ofprobabilities of given table = 0.3 + 0.2 + 0.4 + 0.1 + 0.05
= 1.05 ≠ 1
Hence, the given tableis not the probability distributions of a random variable.
Answer - 2 : -
Given urn containing 5red and 2 black balls.
Let R represent redball and B represent black ball.
Two balls are drawnrandomly.
Hence, the samplespace of the experiment is S = {BB, BR, RB, RR}
X represents thenumber of black balls.
⇒ X (BB) = 2
X (BR) = 1
X (RB) = 1
X (RR) = 0
Therefore, X is afunction on sample space whose range is {0, 1, 2}.
Thus, X is a randomvariable which can take the values 0, 1 or 2.
Question - 3 : - Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?
Answer - 3 : -
Given a coin is tossed6 times.
X represents thedifference between the number of heads and the number of tails.
⇒ X (6H, 0T) =|6-0| = 6
X (5H, 1T) = |5-1| = 4
X (4H, 2T) = |4-2| = 2
X (3H, 3T) = |3-3| = 0
X (2H, 4T) = |2-4| = 2
X (1H, 5T) = |1-5| = 4
X (0H, 6T) = |0-6| = 6
Therefore, X is afunction on sample space whose range is {0, 2, 4, 6}.
Thus, X is a randomvariable which can take the values 0, 2, 4 or 6.
Question - 4 : - Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) Number of tails in the simultaneous tosses of three coins.
(iii) Number of heads in four tosses of a coin.
Answer - 4 : - (i)
Given a coin is tossedtwice.
Hence, the samplespace of the experiment is S = {HH, HT, TH, TT}
X represents thenumber of heads.
⇒ X (HH) = 2
X (HT) = 1
X (TH) = 1
X (TT) = 0
Therefore, X is afunction on sample space whose range is {0, 1, 2}.
Thus, X is a randomvariable which can take the values 0, 1 or 2.
As we know,
P (HH) = P (HT) = P(TH) = P (TT) = 1/4
P (X = 0) = P (TT) =1/4
P (X = 1) = P (HT) + P(TH) = 1/4 + 1/4 = 1/2
P (X = 2) = P (HH) =1/4
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 |
| P (X) | 1/4 | 1/2 | 1/4 |
(ii)
Given three coins aretossed simultaneously. Hence, the sample space of the experiment is S = {HHH,HHT, HTH, THH, TTH, THT, HTT, TTT}
X represents thenumber of tails.
As we see, X is afunction on sample space whose range is {0, 1, 2, 3}.
Thus, X is a randomvariable which can take the values 0, 1, 2 or 3.
P (X = 0) = P(HHH) = 1/8
P (X = 1) = P (HHT) +P (HTH) + P (THH) = 1/8 + 1/8 + 1/8 = 3/8
P (X = 2) = P (HTT) +P (THT) + P (TTH) = 1/8 + 1/8 + 1/8 = 3/8
P (X = 3) = P(TTT) = 1/8
Hence, the required probabilitydistribution is,
| X | 0 | 1 | 2 | 3 |
| P (X) | 1/8 | 3/8 | 3/8 | 1/8 |
(iii)
Given four tosses of acoin.
Hence, the samplespace of the experiment is
S = {HHHH, HHHT, HHTH,HTHH, HTTH, HTHT, HHTT, HTTT, THHH, TTHH, THTH, THHT, THTT, TTHT, TTTH, TTTT}
X represents thenumber of heads.
As we see, X is afunction on sample space whose range is {0, 1, 2, 3, 4}.
Thus, X is a randomvariable which can take the values 0, 1, 2, 3 or 4.
P (X = 0) = P(TTTT) = 1/16
P (X = 1) = P (HTTT) +P (TTTH) + P (THTT) + P (TTHT) = 1/16 + 1/16 + 1/16 + 1/16 = ¼
P(X = 2) = P (HHTT) +P (THHT) + P (TTHH) + P (THTH) + P (HTHT) + P(HTTH)= 1 /16 + 1/16 + 1/16 +1/16 + 1/16 + 1/16 = 6/16 = 3/8
P(X = 3) = P (THHH) +P (HHHT) + P (HTHH) + P (HHTH) = 1/16 + 1/16 + 1/16 + 1/16 = ¼
P(X = 4) = P(HHHH) = 1/16
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 | 3 | 4 |
| P (X) | 1/16 | 1/4 | 3/8 | 1/4 | 1/16 |
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die
Answer - 5 : -
Given a die is tossedtwo times.
When a die is tossedtwo times then the number of observations will be (6 × 6) = 36.
Now, let X is a randomvariable which represents the success.
(i) Here success isgiven as the number greater than 4.
Now
P (X = 0) = P (number≤ 4 in both tosses) = 4/6 × 4/6 = 4/9
P (X = 1) = P (number≤ 4 in first toss and number ≥ 4 in second case) + P (number ≥ 4 in first tossand number ≤ 4 in second case) is
= (4/6 × 2/6) + (2/6 ×4/6) = 4/9
P (X = 2) = P (number≥ 4 in both tosses) = 2/6 × 2/6 = 1/9
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 |
| P (X) | 4/9 | 4/9 | 1/9 |
(ii) Here success isgiven as six appears on at least one die.
Now P (X = 0) = P (sixdoes not appear on any of die) = 5/6 × 5/6 = 25/36
P (X = 1) = P (sixappears at least once of the die) = (1/6 × 5/6) + (5/6 × 1/6) = 10/36 = 5/18
Hence, the requiredprobability distribution is,
Question - 6 : - From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Answer - 6 : -
Given a lot of 30bulbs which include 6 defectives.
Then number ofnon-defective bulbs = 30 – 6 = 24
As 4 bulbs are drawnat random with replacement.
Let X denotes thenumber of defective bulbs from the selected bulbs.
Clearly, X can takethe value of 0, 1, 2, 3 or 4.
P (X = 0) = P (4 arenon-defective and 0 defective)
P (X = 1) = P (3 arenon-defective and 1 defective)
P (X = 2) = P (2 arenon-defective and 2 defective)
P (X = 3) = P (1 arenon-defective and 3 defective)
P (X = 4) = P (0 arenon-defective and 4 defective)
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 | 3 | 4 |
| P (X) | 256/625 | 256/625 | 96/625 | 16/625 | 1/625 |
Question - 7 : - A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
Answer - 7 : -
Given head is 3 timesas likely to occur as tail.
Now, let theprobability of getting a tail in the biased coin be x.
⇒ P (T) = x
And P (H) = 3x
For a biased coin, P(T) + P (H) = 1
⇒ x + 3x = 1
⇒ 4x = 1
⇒ x = 1/4
Hence, P (T) =1/4 and P (H) = 3/4
As the coin is tossedtwice, so the sample space is {HH, HT, TH, TT}
Let X be a randomvariable representing the number of tails.
Clearly, X can takethe value of 0, 1 or 2.
P(X = 0) = P (no tail)= P (H) × P (H) = ¾ × ¾ = 9/16
P(X = 1) = P (onetail) = P (HT) × P (TH) = ¾. ¼ × ¼. ¾ = 3/8
P(X = 2) = P (twotail) = P (T) × P (T) = ¼ × ¼ = 1/16
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 |
| P (x) | 9/16 | 3/8 | 1/16 |
Question - 8 : - A random variable X has the following probability distribution:
Determine
(i) k
(ii) P (X < 3)
(iii) P (X > 6)
(iv) P (0 < X < 3)
Answer - 8 : -
Given a randomvariable X with its probability distribution.
(i) As we know the sumof all the probabilities in a probability distribution of a random variablemust be one.
Hence the sum ofprobabilities of given table:
⇒ 0 + k + 2k + 2k+ 3k + k2 + 2k2 + 7K2 + k = 1
⇒ 10K2 +9k = 1
⇒ 10K2 +9k – 1 = 0
⇒ (10K-1) (k + 1)= 0
k = -1, 1/10
It is known thatprobability of any observation must always be positive that it can’t benegative.
So k = 1/10
(ii) Now we have tofind P(X < 3)
P(X < 3) = P(X = 0)+ P(X = 1) + P(X = 2)
= 0 + k + 2k
= 3k
P (X < 3) = 3 ×1/10 = 3/10
(iii) Now we have findP(X > 6)
P(X > 6) = P(X = 7)
= 7K2 +k
= 7 × (1/10)2 +1/10
= 7/100 + 1/10
P (X > 6) = 17/100
(iv) Consider P (0< X < 3)
P (0 < X < 3) =P(X = 1) + P(X = 2)
= k + 2k
= 3k
P (0 < X < 3) =3 × 1/10 = 3/10
Question - 9 : - The random variable X has a probability distribution P(X) of thefollowing form, where k is some number:
(a) Determine the value of k.
(b) Find P (X < 2), P (X ≤ 2), P(X ≥ 2).
Answer - 9 : -
Given: A randomvariable X with its probability distribution.
(a) As we know the sumof all the probabilities in a probability distribution of a random variablemust be one.
Hence the sum ofprobabilities of given table:
⇒ k + 2k + 3k + 0= 1
⇒ 6k = 1
k = 1/6
(b) Now we have tofind P(X < 2)
P (X < 2) = P (X =0) + P (X = 1)
= k + 2k
= 3k
P (X < 2) = 3 × 1/6= ½
Consider P (X ≤ 2)
P (X ≤ 2) = P (X = 0)+ P (X = 1) + P (X = 2)
= k + 2k + 3k
= 6k
P (X ≤ 2) = 6 × 1/6 =1
Now we have to findP(X ≥ 2)
P(X ≥ 2) = P(X = 2) +P(X > 2)
= 3k + 0
= 3k
P (X ≥ 2) = 3 × 1/6 =½
Question - 10 : - Find the mean number of heads in three tosses of a fair coin.
Answer - 10 : -
Given a coin is tossedthree times.
Three coins are tossedsimultaneously. Hence, the sample space of the experiment is S = {HHH, HHT,HTH, THH, TTH, THT, HTT, TTT}
X represents thenumber of heads.
As we see, X is afunction on sample space whose range is {0, 1, 2, 3}.
Thus, X is a randomvariable which can take the values 0, 1, 2 or 3.
P (X = 0) = P(TTT) = 1/8
P (X = 1) = P (TTH) +P (THT) + P (HTT) = 1/8 +1/8 + 1/8 = 3/8
P (X = 2) = P (THH) +P (HTH) + P (HHT) = 1/8 + 1/8 + 1/8 = 3/8
P (X = 3) = P (HHH) =1/8
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 | 3 |
| P (X) | 1/8 | 3/8 | 3/8 | 1/8 |
Therefore mean μ is:
