RD Sharma Solutions for Class 12 Maths Exercise 33.1 Chapter 33 Binomial Distribution are available here. This exercise of RD Sharma Solutions for Class 12 Maths Chapter 33 contains topics related to Probability. To cover the entire syllabus in Maths, the RD Sharma is an essential material, as it offers a wide range of questions that test students’ understanding of concepts.
RD Sharma Solution for Class 12 Maths Chapter 33 Exercise 1
Access Answers for RD Sharma Solution Class 12 Maths Chapter 33 Exercise 1
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.
Comments