EXERCISE 33.1
Question. 1
Solution:
Let us assume P denotes the probability of having a defective item.
From the question, P = 6%
= 6/100
P = 3/50
We know that, P + Q = 1
Then, Q = 1 – P
Q = 1 – 3/50
Q = (50 – 3)/50
Q = 47/50
Now assume X denotes the number of defective items in a sample of 8 items.
So, probability of getting r defective bulks is P(X – r) = nCrPrQn – r
P(X – r) = 8Cr(3/50)r (47/50)8 – r … [equation (i)]
Hence, the probability of getting not more than one defective item
= P(X = 0) + P(X = 1)
By using equation (i)
= 8C0(3/50)0 (47/50)8-0 + 8C1 (3/50)1 (47/50)8-1
= 1.1(47/50)8 + 8 (3/50) (47/50)7
= (47/50)7 (47/50 + 24/50)
= (71/50) (47/50)7
= (1.42) (0.94)7
Therefore, the required probability is (1.42) (0.94)7.
Question. 2
Solution:
From the question, it is given that,
A coin is tossed 5 times.
So, the probability of getting head on one throw of coin = ½
P = ½
We know that, P + Q = 1
Q = 1 – ½
Q = (2 – 1)/2
Q = ½
Now assume X denotes the number of getting heads as 5 tosses of coins.
So, the probability of getting r heads in n tosses of the coin is given by P(X – r) = nCrPrQn – r
P(X – r) = 5Cr(1/2)r (1/2)5 – r … [equation (i)]
Hence, the probability of getting at least 3 heads
= P(X = 3) + P(X = 4) + P(X = 5)
By using equation (i)
= 5C3(½)3 (½)5-3 + 5C4 (½)4 (½)5-4 + 5C5 (½)5 (½)0
= 5C3(½)3 (½)2 + 5C4 (½)4 (½) + 5C5 (½)5 (1)
= (5 × 4)/2 (½)5 + 5 (½)5 + 1 (½)5
= (½)5 (10 + 5 + 1)
= 16 (1/32)
= 16/32
= ½
Therefore, the required probability is ½.
Question. 3
Solution:
Let us assume P denotes the probability of getting tail on a toss of a fair coin,
We know that, P + Q = 1
Then, Q = 1 – P
Q = 1 – ½
Q = (2 – 1)/2
Q = 1/2
Now assume X denotes the number of tails obtained on the toss of a coin 5 times.
So, the probability of getting r tails in n tosses of the coin is given by P(X – r) = nCrPrQn – r
P(X – r) = 5Cr (½)r (½)5 – r … [equation (i)]
Hence, the probability of getting the tail an odd number of times
= P(X = 1) + P(X = 3) + P(x = 5)
By using equation (i)
= 5C3(½)1 (½)5-1 + 5C4 (½)3 (½)5-3 + 5C5 (½)5 (½)0
= 5C3(½)1 (½)4 + 5C4 (½)3 (½)2 + 5C5 (½)5 (1)
= (5 × 4)/2 (½)5 + 5 (½)5 + 1 (½)5
= (½)5 (5 + 10 + 1)
= 16 (½)5
= 16(1/32)
= 16/32
= ½
Therefore, the required probability is ½.
Question. 4
Solution:
Let us assume P be the probability of getting a sum of 9, and it is considered a success.
From the question, it is given that a pair of dice is thrown 6 times,
= [(3, 6), (4, 5), (5, 4), (6, 3)]
Then, P = 4/36 … [divide both sides by 4]
P = 1/9
We know that, P + Q = 1
Then, Q = 1 – P
Q = 1 – 1/9
Q = (9 – 1)/9
Q = 8/9
Now assume X be the number of successes in three of a pair of dice 6 times.
So, the probability of getting r success out of n is given by
P(X – r) = nCrPrQn – r … [equation (i)]
Hence, the probability of getting at least 5 successes,
= P(X = 5) + P(X = 6)
By using equation (i)
= 6C5 (1/9)5 (8/9)6-5 + 6C6 (1/9)6 (8/9)6-6
= 6 (1/9)5 (8/9) + 1 (1/9)6 (8/9)0
= (1/9)5 (48/9 + 1/9)
= (49/9) (1/9)5
= 49/96
Therefore, the required probability is 49/96.
Question. 5
Solution:
Let us assume P be the probability of getting head in a throw of a coin.
Then, P = ½
We know that, P + Q = 1
Then, Q = 1 – P
Q = 1 – 1/2
Q = (2 – 1)/2
Q = ½
Now assume X be the number of heads on tossing the coin 6 times.
So, the probability of getting r tossing the coin n times is given by
P(X – r) = nCrPrQn – r … [equation (i)]
Hence, the probability of getting at least 3 heads,
= P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
= 1 – (P(X = 0) + P(X = 1) + P(X = 2))
By using equation (i)
= 1 – (6C0 (½)0 (½)6-0 + 6C1 (½)6 (½)6-1 + 6C2 (½)2 (½)6-2)
= 1 – (1 (½)6 + 6 (½)6 + 6.5/2 (½)6)
= 1 – ((½)6 (1 + 6 + 15))
= 1 – (22/64)
= (64 – 22)/64
= 42/64
= 21/32
Therefore, the required probability is 21/32.
EXERCISE 33.2
Question. 1
Solution:
Now, let us assume x be a binomial variate with parameters n and p,
So, Mean = np
and Variance = npq
Then, Mean – Variance = np – npq
= np(1 – q)
= np(p)
= np2
Mean – Variance > 0
Mean > Variance
Therefore, mean can never be less than the variance.
Question. 2
Solution:
From the question, it is given that mean = 9 and variance = 9/4
Now, let us assume x be a binomial variate with parameters n and p,
P + Q = 1
Q = 1 – P
So, Mean = np = 9 … [equation (i)]
Variance = npq = 9/4 … [equation (ii)]
By dividing equation (i) by equation (ii), we get,
npq/np = (9/4)/9
q = ¼
Then, P = 1 – Q
Substituting the value of Q, we get,
P = 1 – ¼
P = (4 – 1)/4
P = ¾
Now, substitute the value of p in equation (i), AND we get,
n (¾) = 9
n = 36/3
n = 12
Then, the distribution is given by = nCrPrQn – r
P(X = r) = 12Cr (¾)r (¼)12 – r
Therefore, r = 0, 1, 2, …. 12
Question. 3
Solution:
From the question, it is given that mean = 9 and variance = 6
Now, let us assume x be a binomial variate with parameters n and p,
P + Q = 1
Q = 1 – P
So, Mean = np = 9 … [equation (i)]
Variance = npq = 6 … [equation (ii)]
By dividing equation (i) by equation (ii), we get,
npq/np = 6/9
q = 2/3
Then, P = 1 – Q
Substituting the value of Q, we get,
P = 1 – 2/3
P = (3 – 2)/3
P = 1/3
Now, substitute the value of p in equation (i) we get,
n (1/3) = 9
n = 27
Then, the distribution is given by = nCrPrQn – r
P(X = r) = 27Cr (1/3)r (2/3)27 – r
Therefore, r = 0, 1, 2, …. 27
Question. 4
Solution:
From the question, it is given that the sum of mean and variance for 5 trials is 4.8
So, n = 5
Then, Mean + Variance = 4.8
np + npq = 4.8
By taking out common terms, we get,
np(1 + q) = 4.8
Substitute the value of n,
5p (1 + q) = 4.8
We know that, P + Q = 1
5 (1 + q) (1 – q) = 4.8
5(1 – q2) = 4.8
1 – q2 = 4.8/5
q2 = 1 – 4.8/5
q2 = (5 – 4.8)/5
q2 = 0.2/5
q2 = 2/50
q2 = 1/25
q = √(1/25)
q = 1/5
Then, P = 1- q
P = 1 – 1/5
P = (5 -1)/5
P = 4/5
Therefore, n = 5, p = 4/5 and q = 1/5
So, the binomial distribution is
P(X = r) = nCrPrQn – r
P(X = r) = = 5Cr(4/5)r (1/5)5 – r
r = 0, 1, 2, 3, … 5
Question. 5
Solution:
From the question, it is given that mean = 20 and variance = 16
Now, let us assume parameters n and p of distribution,
So, Mean = np = 20 … [equation (i)]
Variance = npq = 16 … [equation (ii)]
By dividing equation (i) by equation (ii), we get,
npq/np = 16/20
q = 4/5
We know that, P + Q = 1
Then, P = 1 – Q
Substituting the value of Q, we get,
P = 1 – 4/5
P = (5 – 4)/5
P = 1/5
Now, substitute the value of p in equation (i), and we get,
n (1/5) = 20
n = 100
Then, the distribution is given by = nCrPrQn – r
P(X = r) = 100Cr (1/5)r (4/5)100 – r
Therefore, r = 0, 1, 2, …. 100
Question. 6
Solution:
From the question, it is given that the sum and product of the mean and the variance are 25/3 and 50/3, respectively.
Now, let us assume parameters n and p of the binomial distribution.
We know that, P + Q = 1
Q = 1 – P
Then, Mean + Variance = 25/3
np + npq = 25/3
By taking common terms outside, we get,
np (1 + q) = 25/3
np = 25/3(1 + q) … [equation (i)]
Now, Mean × Variance = 50/3
np × npq = 50/3
n2p2q = 50/q
From equation (i)
(25/3(1 + q))2 q = 50/3
625q = 50/3 (9(1 + q)2)
625q = 150 (1 + q)2
25q = 6(1 + q)2
6 + 6q2 + 12q – 25q = 0
6q2 – 13q + 6 = 0
6q2 – 9q – 4q + 6 = 0
By taking common terms outside, we get,
3q (2q – 3) – 2 (2q – 3) = 0
(2q – 3) (3q – 2) = 0
2q – 3 = 0, 3q – 2 = 0
q = 3/2, q = 2/3
Since q ≤ 1
q = 2/3
P = 1 – q
Substitute the value of q,
P = 1 – (2/3)
P = (3 – 2)/3
P = 1/3
Now, substitute the value of p and q in equation (i) we get,
np = 25/3(1 + q)
By cross multiplication,
3np (1 + q) = 25
3n (1/3) (1 + 2/3) = 25
3n (1/3) ((3 + 2)/3) = 25
3n (1/3) (5/3) = 25
n (5/3) = 25
n = 75/5
n = 15
Then, the binomial distribution is given by = nCrPrQn – r
P(X = r) = 15Cr (1/3)r (2/3)15 – r
Therefore, r = 0, 1, 2, …. 15