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Question
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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.
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One school is chosen at random. Then,
- find the probability that the school chosen at random has more than 100 computers.
- find the probability that the school chosen at random has 50 or fewer computers.
- find the probability that the school chosen at random has no more than 20 computers.
- find the probability that the school chosen at random has 10 or less than 10 computers.
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Solution
Given Data:
Total number of surveyed schools: n(S) = 1000
i. Probability that the chosen school has more than 100 computers:
Favourable category: '101 and more'
Number of favourable schools: \[n(E_1) = 80\]
Formula: \[P(E_1) = \frac{n(E_1)}{n(S)}\]
Calculation: \[P(\text{more than } 100) = \frac{80}{1000} = \frac{2}{25}\]
ii. Probability that the chosen school has 50 or fewer computers:
Favourable categories: '1–10', '11–20' and '21–50'
Number of favourable schools: \[n(E_2) = 250 + 200 + 290 = 740\]
Formula: \[P(E_2) = \frac{n(E_2)}{n(S)}\]
Calculation: \[P(\le 50) = \frac{740}{1000} = \frac{37}{50}\]
iii. Probability that the chosen school has no more than 20 computers:
'No more than 20' means at most 20 computers, i.e., categories '1–10' and '11–20'.
Number of favourable schools: \[n(E_3) = 250 + 200 = 450\]
Formula: \[P(E_3) = \frac{n(E_3)}{n(S)}\]
Calculation: \[P(\le 20) = \frac{450}{1000} = \frac{9}{20}\]
iv. Probability that the chosen school has 10 or less than 10 computers:
Favourable category: '1–10'
Number of favourable schools: \[n(E_4) = 250\]
Formula: \[P(E_4) = \frac{n(E_4)}{n(S)}\]
Calculation: \[P(\le 10) = \frac{250}{1000} = \frac{1}{4}\]
