RD Chapter 33 Probability Ex 33.3 Solutions
Question - 21 : - A box contains 100bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that:
(i) all 10 are defective
(ii) all 10 are good
(iii) at least one is defective
(iv) none is defective
Answer - 21 : -
Given: A box contains100bulbs, 20 of which are defective.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
Ten bulbs are drawn atrandom for inspection,
Total possibleoutcomes are 100C10
n (S) = 100C10
(i) Let E be theevent that all ten bulbs are defective
n (E) = 20C10
P (E) = n (E) / n (S)
= 20C10 / 100C10
(ii) Let E be theevent that all ten good bulbs are selected
n (E) = 80C10
P (E) = n (E) / n (S)
= 80C10 / 100C10
(iii) Let E be theevent that at least one bulb is defective
E={1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defectivebulbs
Let E′ be the eventthat none of the bulb is defective
n (E′) = 80C10
P (E′) = n (E′) / n(S)
= 80C10 / 100C10
So, P (E) = 1 – P (E′)
= 1 – 80C10 / 100C10
(iv) Let E be theevent that none of the selected bulb is defective
n (E) = 80C10
P (E) = n (E) / n (S)
= 80C10 / 100C10
Question - 22 : - Find the probability that in a random arrangement of the letters of the word ‘SOCIAL’ vowels come together
Answer - 22 : -
Given: The word‘SOCIAL’.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
In the randomarrangement of the alphabets of word “SOCIAL” we have to find the probabilitythat vowels come together.
Total possibleoutcomes of arranging the alphabets are 6!
n (S) = 6!
Let E be the eventthat vowels come together
Number of vowels inSOCIAL is A, I, O
So, number of ways toarrange them where, (A, I, O) come together
n (E) = 4! × 3!
P (E) = n (E) / n (S)
= [4! × 3!] / 6!
= 1/5
Question - 23 : - The letters of the word ‘CLIFTON’ are placed at random in a row. What is the chance that two vowels come together?
Answer - 23 : -
Given: The word‘CLIFTON’.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
In the randomarrangement of the alphabets of word “CLIFTON” we have to find the probabilitythat vowels come together.
Total possibleoutcomes of arranging the alphabets are 7!
n (S) =7!
Let E be the eventthat vowels come together
Number of vowels inCLIFTON is I, O
Number of ways toarrange them where, (I, O) come together
n (E)= 6! × 2!
P (E) = n (E) / n (S)
= [6! × 2!] / 7!
= 2/7