MathematicsQ51–57 of 57 questions
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51
MathematicsEasyMCQGSEB · 2022
Value of the objective function $Z = -50x + 20y$ subject to the constraints $2x - y \geq -5$,$3x + y \geq 3$,$2x - 3y \leq 12$,$x \geq 0$,$y \geq 0$. The corner points of the feasible region are $(0, 5)$,$(0, 3)$,$(1, 0)$,and $(6, 0)$. At which point is the value of $Z$ minimum?
A
$(0, 3)$
B
$(6, 0)$
C
$(0, 5)$
D
$(1, 0)$
Solution
(B) To find the minimum value of the objective function $Z = -50x + 20y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At $(0, 5)$: $Z = -50(0) + 20(5) = 100$
$2$. At $(0, 3)$: $Z = -50(0) + 20(3) = 60$
$3$. At $(1, 0)$: $Z = -50(1) + 20(0) = -50$
$4$. At $(6, 0)$: $Z = -50(6) + 20(0) = -300$
Comparing the values $100, 60, -50$,and $-300$,the minimum value is $-300$,which occurs at the point $(6, 0)$.
$1$. At $(0, 5)$: $Z = -50(0) + 20(5) = 100$
$2$. At $(0, 3)$: $Z = -50(0) + 20(3) = 60$
$3$. At $(1, 0)$: $Z = -50(1) + 20(0) = -50$
$4$. At $(6, 0)$: $Z = -50(6) + 20(0) = -300$
Comparing the values $100, 60, -50$,and $-300$,the minimum value is $-300$,which occurs at the point $(6, 0)$.
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52
MathematicsEasyMCQGSEB · 2022
The corner points of the feasible region determined by the system of linear constraints are $(2, 72)$,$(15, 20)$,and $(40, 15)$. Let $Z = 6x + 3y$ be the objective function. The minimum value of $Z$ occurs at:
A
$(15, 20)$
B
$(2, 72)$
C
$(40, 15)$
D
$(0, 11)$
Solution
(A) To find the minimum value of the objective function $Z = 6x + 3y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At point $(2, 72)$: $Z = 6(2) + 3(72) = 12 + 216 = 228$
$2$. At point $(15, 20)$: $Z = 6(15) + 3(20) = 90 + 60 = 150$
$3$. At point $(40, 15)$: $Z = 6(40) + 3(15) = 240 + 45 = 285$
Comparing the values $228$,$150$,and $285$,the minimum value is $150$,which occurs at the point $(15, 20)$.
$1$. At point $(2, 72)$: $Z = 6(2) + 3(72) = 12 + 216 = 228$
$2$. At point $(15, 20)$: $Z = 6(15) + 3(20) = 90 + 60 = 150$
$3$. At point $(40, 15)$: $Z = 6(40) + 3(15) = 240 + 45 = 285$
Comparing the values $228$,$150$,and $285$,the minimum value is $150$,which occurs at the point $(15, 20)$.
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53
MathematicsEasyMCQGSEB · 2022
The probability that a student is not a swimmer is $\frac{1}{5}$. Then,the probability that out of $5$ students,$4$ are swimmers is . . . . . . .
A
$^5C_4 \left(\frac{4}{5}\right)^4 \left(\frac{1}{5}\right)^1$
B
$\left(\frac{4}{5}\right)^4 \frac{1}{5}$
C
$5\left(\frac{4}{5}\right)^4$
D
None of these
Solution
(D) Let $p$ be the probability that a student is a swimmer and $q$ be the probability that a student is not a swimmer.
Given $q = \frac{1}{5}$,so $p = 1 - q = 1 - \frac{1}{5} = \frac{4}{5}$.
We need to find the probability that out of $n = 5$ students,$x = 4$ are swimmers.
Using the Binomial Distribution formula: $P(X = x) = ^nC_x \cdot p^x \cdot q^{n-x}$.
Substituting the values: $P(X = 4) = ^5C_4 \cdot \left(\frac{4}{5}\right)^4 \cdot \left(\frac{1}{5}\right)^{5-4}$.
$P(X = 4) = 5 \cdot \left(\frac{4}{5}\right)^4 \cdot \frac{1}{5} = \left(\frac{4}{5}\right)^4$.
Since this value is not explicitly listed in the options $A, B, C$,the correct choice is $D$.
Given $q = \frac{1}{5}$,so $p = 1 - q = 1 - \frac{1}{5} = \frac{4}{5}$.
We need to find the probability that out of $n = 5$ students,$x = 4$ are swimmers.
Using the Binomial Distribution formula: $P(X = x) = ^nC_x \cdot p^x \cdot q^{n-x}$.
Substituting the values: $P(X = 4) = ^5C_4 \cdot \left(\frac{4}{5}\right)^4 \cdot \left(\frac{1}{5}\right)^{5-4}$.
$P(X = 4) = 5 \cdot \left(\frac{4}{5}\right)^4 \cdot \frac{1}{5} = \left(\frac{4}{5}\right)^4$.
Since this value is not explicitly listed in the options $A, B, C$,the correct choice is $D$.
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54
MathematicsEasyMCQGSEB · 2022
If $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$,then which of the following is correct?
A
$P(A \mid B) = \frac{P(B)}{P(A)}$
B
$P(A \mid B) < P(A)$
C
$P(A \mid B) \geq P(A)$
D
None of these
Solution
(C) Given that $A \subset B$,it implies that $A \cap B = A$.
By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting $A \cap B = A$,we get $P(A \mid B) = \frac{P(A)}{P(B)}$.
Since $A \subset B$,we have $P(A) \leq P(B)$,which implies $\frac{1}{P(B)} \leq \frac{1}{P(A)}$.
Multiplying both sides by $P(A)$ (assuming $P(A) > 0$),we get $\frac{P(A)}{P(B)} \leq 1$.
Also,since $P(B) \leq 1$,we have $\frac{P(A)}{P(B)} \geq P(A)$.
Thus,$P(A \mid B) \geq P(A)$.
By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting $A \cap B = A$,we get $P(A \mid B) = \frac{P(A)}{P(B)}$.
Since $A \subset B$,we have $P(A) \leq P(B)$,which implies $\frac{1}{P(B)} \leq \frac{1}{P(A)}$.
Multiplying both sides by $P(A)$ (assuming $P(A) > 0$),we get $\frac{P(A)}{P(B)} \leq 1$.
Also,since $P(B) \leq 1$,we have $\frac{P(A)}{P(B)} \geq P(A)$.
Thus,$P(A \mid B) \geq P(A)$.
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55
MathematicsEasyMCQGSEB · 2022
For two events $A$ and $B$,$P(B) \neq 0$ and $P(A \mid B) = 1$,then which of the following is true?
A
$A = B$
B
$B \subset A$
C
$A \subset B$
D
$A = \phi$
Solution
(B) By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Given $P(A \mid B) = 1$,we have $\frac{P(A \cap B)}{P(B)} = 1$,which implies $P(A \cap B) = P(B)$.
Since $A \cap B$ is a subset of $B$ $(A \cap B \subseteq B)$,and their probabilities are equal,it follows that all outcomes in $B$ must be contained within $A$.
Therefore,$B \subseteq A$ (or $B \subset A$ in the context of the given options).
Given $P(A \mid B) = 1$,we have $\frac{P(A \cap B)}{P(B)} = 1$,which implies $P(A \cap B) = P(B)$.
Since $A \cap B$ is a subset of $B$ $(A \cap B \subseteq B)$,and their probabilities are equal,it follows that all outcomes in $B$ must be contained within $A$.
Therefore,$B \subseteq A$ (or $B \subset A$ in the context of the given options).
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56
MathematicsEasyMCQGSEB · 2022
For independent events $A$ and $B$,$P(A) = \frac{1}{3}$,$P(A \cup B) = \frac{2}{5}$,and $P(B) = p$. Find the value of $p$.
A
$\frac{1}{20}$
B
$\frac{2}{5}$
C
$\frac{1}{10}$
D
$\frac{1}{5}$
Solution
(C) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \cdot P(B)$.
We know the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{2}{5} = \frac{1}{3} + p - (\frac{1}{3} \cdot p)$.
$\frac{2}{5} = \frac{1}{3} + p(1 - \frac{1}{3})$.
$\frac{2}{5} = \frac{1}{3} + p(\frac{2}{3})$.
Subtracting $\frac{1}{3}$ from both sides: $\frac{2}{5} - \frac{1}{3} = p(\frac{2}{3})$.
$\frac{6 - 5}{15} = p(\frac{2}{3})$.
$\frac{1}{15} = p(\frac{2}{3})$.
$p = \frac{1}{15} \cdot \frac{3}{2} = \frac{3}{30} = \frac{1}{10}$.
Thus,the correct option is $C$.
We know the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{2}{5} = \frac{1}{3} + p - (\frac{1}{3} \cdot p)$.
$\frac{2}{5} = \frac{1}{3} + p(1 - \frac{1}{3})$.
$\frac{2}{5} = \frac{1}{3} + p(\frac{2}{3})$.
Subtracting $\frac{1}{3}$ from both sides: $\frac{2}{5} - \frac{1}{3} = p(\frac{2}{3})$.
$\frac{6 - 5}{15} = p(\frac{2}{3})$.
$\frac{1}{15} = p(\frac{2}{3})$.
$p = \frac{1}{15} \cdot \frac{3}{2} = \frac{3}{30} = \frac{1}{10}$.
Thus,the correct option is $C$.
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57
MathematicsEasyMCQGSEB · 2022
$3 P(A) = P(B) = \frac{5}{13}$ and $P(A \mid B) = \frac{3}{5}$,then $P(A \cup B) = $ . . . . . . .
A
$\frac{15}{39}$
B
$\frac{11}{39}$
C
$\frac{17}{39}$
D
$\frac{13}{39}$
Solution
(B) Given: $3 P(A) = \frac{5}{13} \implies P(A) = \frac{5}{39}$.
$P(B) = \frac{5}{13} = \frac{15}{39}$.
$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{3}{5}$.
Therefore,$P(A \cap B) = P(A \mid B) \times P(B) = \frac{3}{5} \times \frac{5}{13} = \frac{3}{13} = \frac{9}{39}$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{5}{39} + \frac{15}{39} - \frac{9}{39} = \frac{11}{39}$.
$P(B) = \frac{5}{13} = \frac{15}{39}$.
$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{3}{5}$.
Therefore,$P(A \cap B) = P(A \mid B) \times P(B) = \frac{3}{5} \times \frac{5}{13} = \frac{3}{13} = \frac{9}{39}$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{5}{39} + \frac{15}{39} - \frac{9}{39} = \frac{11}{39}$.
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