In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Acc...
Question 1 Report
In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Accounts only, 2 Accounts and Economics only. If each student offered at least one of the subjects,
(a) how many students failed in all subjects?
(b) find the percentage number that failed in at least one of Economics and Mathematics
(c) calculate the probability that a student picked at random failed in Accounts?
Answer Details
Let the regions of the three-set Venn diagram (M = Mathematics, A = Accounts, E = Economics) be filled from the "only" data. Given: \(M\) only \(=6\), \(A\) only \(=9\), \(M\cap A\) only \(=5\), \(A\cap E\) only \(=2\).
Let the triple region \(M\cap A\cap E=t\).
Accounts total \(=19\): \(9+5+2+t=19\Rightarrow t=3\).
Mathematics total \(=18\): \(6+5+t+(M\cap E\ \text{only})=18\Rightarrow 6+5+3+(M\cap E\ \text{only})=18\Rightarrow M\cap E\ \text{only}=4\).
Economics total \(=16\): \((E\ \text{only})+4+2+3=16\Rightarrow E\ \text{only}=7\).
Sum of all seven regions \(=6+9+7+5+4+2+3=36\).
(a) Since each of the 40 students offered at least one subject, those in no region failed all three:
\[40-36=4\text{ students failed all subjects.}\](b) Failed at least one of Economics and Mathematics. The complement is passing BOTH Mathematics and Economics \(=(M\cap E\ \text{only})+t=4+3=7\). So the number failing at least one of them \(=40-7=33\).
\[\frac{33}{40}\times100\%=82.5\%\](c) Probability of failing Accounts. Number passing Accounts \(=19\), so number failing Accounts \(=40-19=21\).
\[P(\text{failed Accounts})=\frac{21}{40}\]Download The App On Google Playstore
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