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Case Study - 3
Computer-based learning (CBL) refers to any teaching methodology that makes use of computers for information transmission. At an elementary school level, computer applications can be used to display multimedia lesson plans. A survey was done on $1000$ elementary and secondary schools of Assam and they were classified by the number of computers they had.
Number of Computers
Number of Schools
1-10
250
11-20
200
21-50
290
51-100
180
101 and more
80
One school is chosen at random. Then :
(i) Find the probability that the school chosen at random has more than $100$ computers.
(ii) (a) Find the probability that the school chosen at random has $50$ or fewer computers.
OR
(ii) (b) Find the probability that the school chosen at random has no more than $20$ computers.
(iii) Find the probability that the school chosen at random has $10$ or less than $10$ computers.
Computer-based learning (CBL) refers to any teaching methodology that makes use of computers for information transmission. At an elementary school level, computer applications can be used to display multimedia lesson plans. A survey was done on $1000$ elementary and secondary schools of Assam and they were classified by the number of computers they had.
Number of Computers
Number of Schools
1-10
250
11-20
200
21-50
290
51-100
180
101 and more
80
One school is chosen at random. Then :
(i) Find the probability that the school chosen at random has more than $100$ computers.
(ii) (a) Find the probability that the school chosen at random has $50$ or fewer computers.
OR
(ii) (b) Find the probability that the school chosen at random has no more than $20$ computers.
(iii) Find the probability that the school chosen at random has $10$ or less than $10$ computers.
Show SolutionHide Solution↓
(i) P (more than $100$ computers) = $\frac{80}{1000}$ or $0.08$
(ii)(a) $50$ or fewer computers = $250 + 200 + 290 = 740$
Required probability = $\frac{740}{1000}$ or $0.74$
OR
(ii)(b) No more than $20$ computers = $250 + 200 = 450$
Required probability = $\frac{450}{1000}$ or $0.45$
(iii) P ($10$ or less than $10$ computer) = $\frac{250}{1000}$ or $0.25$
(ii)(a) $50$ or fewer computers = $250 + 200 + 290 = 740$
Required probability = $\frac{740}{1000}$ or $0.74$
OR
(ii)(b) No more than $20$ computers = $250 + 200 = 450$
Required probability = $\frac{450}{1000}$ or $0.45$
(iii) P ($10$ or less than $10$ computer) = $\frac{250}{1000}$ or $0.25$