Question

A student takes his examination in four subjects $\alpha ,\beta ,\gamma ,\delta$. He estimates his chance of passing in $\alpha$ is $\dfrac{4}{5}$ , $\beta$ is $\dfrac{3}{4}$ , in $\gamma$ is $\dfrac{5}{6}$ and $\delta$ is $\dfrac{2}{3}$ . The probability that he qualifies (passes in at least three subjects) is
A. $\dfrac{{34}}{{90}}$
B. $\dfrac{{61}}{{90}}$
C. $\dfrac{{53}}{{90}}$
D. None of these

Answer 1

Hint : In this word problem mentioned as a student takes four subjects in his examination. And here also estimated his chance of passing in four subjects. So, we need to find the possibilities of passes in at least three subjects. We need the probability formula is ${\rm P}(\overline {\rm A} ) = 1 - {\rm P}({\rm A})$

Complete step by step solution:
Let as assume, the given possibility of qualifying the examination in four subject, we have

$$\alpha = \dfrac{4}{5}, \\\ \beta = \dfrac{3}{4}, \\\ \gamma = \dfrac{5}{6}, \\\ \delta = \dfrac{2}{3} \;$$

By substituting the above values into the probability formula, ${\rm P}(\overline {\rm A} ) = 1 - {\rm P}({\rm A})$
The probability of impossible event,
${\rm P}(\overline \alpha ) = 1 - {\rm P}(\alpha ) = 1 - \dfrac{4}{5} = \dfrac{{5 - 4}}{5} = \dfrac{1}{5}$
${\rm P}(\overline \beta ) = 1 - {\rm P}(\beta ) = 1 - \dfrac{3}{4} = \dfrac{{4 - 3}}{4} = \dfrac{1}{4}$
${\rm P}(\overline \gamma ) = 1 - {\rm P}(\gamma ) = 1 - \dfrac{5}{6} = \dfrac{{6 - 5}}{6} = \dfrac{1}{6}$
${\rm P}(\overline \delta ) = 1 - {\rm P}(\delta ) = 1 - \dfrac{2}{3} = \dfrac{{3 - 2}}{3} = \dfrac{1}{3}$
To calculate the different possibilities to qualify are
Possibility of qualify in all the four subject, ${\rm P}(\alpha \beta \gamma \delta )$
Possibility of qualify in three subject out of four, ${\rm P}(\overline \alpha \beta \gamma \delta ),{\rm P}(\alpha \overline \beta \gamma \delta ),{\rm P}(\alpha \beta \overline \gamma \delta ),{\rm P}(\alpha \beta \gamma \overline \delta )$
The probability of impossible event, we have
${\rm P}(\alpha \beta \gamma \delta \cup \overline \alpha \beta \gamma \delta \cup \alpha \overline \beta \gamma \delta \cup \alpha \beta \overline \gamma \delta \cup \alpha \beta \gamma \overline \delta ) = {\rm P}(\alpha \beta \gamma \delta ) + {\rm P}(\overline \alpha \beta \gamma \delta ) + {\rm P}(\alpha \overline \beta \gamma \delta ) + {\rm P}(\alpha \beta \overline \gamma \delta ) + {\rm P}(\alpha \beta \gamma \overline \delta )$
By substitute the values in the probability formula, we get
Qualify in all the four subject, ${\rm P}(\alpha \beta \gamma \delta ) = \dfrac{4}{5} \times \dfrac{3}{4} \times \dfrac{5}{6} \times \dfrac{2}{3} = \dfrac{{120}}{{360}} = \dfrac{1}{3}$
Qualify in three subjects, $\alpha ,\beta ,\gamma$ , not $\delta$ , ${\rm P}(\alpha \beta \gamma \overline \delta ) = \dfrac{4}{5} \times \dfrac{3}{4} \times \dfrac{5}{6} \times \dfrac{1}{3} = \dfrac{{60}}{{360}} = \dfrac{1}{6}$
Qualify in three subjects, $\alpha ,\beta ,\delta$ , not $\gamma$ , ${\rm P}(\alpha \beta \overline \gamma \delta ) = \dfrac{4}{5} \times \dfrac{3}{4} \times \dfrac{1}{6} \times \dfrac{2}{3} = \dfrac{{24}}{{360}} = \dfrac{1}{{15}}$
Qualify in three subjects, $\alpha ,\gamma ,\delta$ , not $\beta$ , ${\rm P}(\alpha \overline \beta \gamma \delta ) = \dfrac{4}{5} \times \dfrac{1}{4} \times \dfrac{5}{6} \times \dfrac{2}{3} = \dfrac{{40}}{{360}} = \dfrac{1}{9}$
Qualify in three subjects, $\beta ,\gamma ,\delta$ ,not $\alpha$ , ${\rm P}(\overline \alpha \beta \gamma \delta ) = \dfrac{1}{5} \times \dfrac{3}{4} \times \dfrac{5}{6} \times \dfrac{2}{3} = \dfrac{{30}}{{360}} = \dfrac{1}{{12}}$
By applying all above value into the probability of impossible event formula,

$${\rm P}(\alpha \beta \gamma \delta + \overline \alpha \beta \gamma \delta \cup \alpha \overline \beta \gamma \delta \cup \alpha \beta \overline \gamma \delta \cup \alpha \beta \gamma \overline \delta ) = {\rm P}(\alpha \beta \gamma \delta ) + {\rm P}(\overline \alpha \beta \gamma \delta ) + {\rm P}(\alpha \overline \beta \gamma \delta ) + {\rm P}(\alpha \beta \overline \gamma \delta ) + {\rm P}(\alpha \beta \gamma \overline \delta ) \\\ {\rm P}(\alpha \beta \gamma \delta \cup \overline \alpha \beta \gamma \delta \cup \alpha \overline \beta \gamma \delta \cup \alpha \beta \overline \gamma \delta \cup \alpha \beta \gamma \overline \delta ) = \dfrac{1}{3} + \dfrac{1}{{12}} + \dfrac{1}{9} + \dfrac{1}{{15}} + \dfrac{1}{6} \;$$

Take LCM on the above equation, we get
${\rm P}(\alpha \beta \gamma \delta + \overline \alpha \beta \gamma \delta \cup \alpha \overline \beta \gamma \delta \cup \alpha \beta \overline \gamma \delta \cup \alpha \beta \gamma \overline \delta ) = \dfrac{{60 + 12 + 20 + 15 + 30}}{{180}}$
To simplify, we get
${\rm P}(\alpha \beta \gamma \delta + \overline \alpha \beta \gamma \delta \cup \alpha \overline \beta \gamma \delta \cup \alpha \beta \overline \gamma \delta \cup \alpha \beta \gamma \overline \delta ) = \dfrac{{137}}{{180}}$
Therefore, the probability of qualifying at least three subjects is $\dfrac{{137}}{{180}}$ .
So, The final answer is Option (D) - None of these.
So, the correct answer is “Option D”.

Note : In this problem, To calculate the possibilities of passes in at least three subjects by the probability formula. We need to remember this concept to solve this type of problem. ${\rm P}(\alpha \beta \gamma \delta \cup \overline \alpha \beta \gamma \delta \cup \alpha \overline \beta \gamma \delta \cup \alpha \beta \overline \gamma \delta \cup \alpha \beta \gamma \overline \delta ) = {\rm P}(\alpha \beta \gamma \delta ) + {\rm P}(\overline \alpha \beta \gamma \delta ) + {\rm P}(\alpha \overline \beta \gamma \delta ) + {\rm P}(\alpha \beta \overline \gamma \delta ) + {\rm P}(\alpha \beta \gamma \overline \delta )$ Here, we need to solve the probability of impossible events.