Aptitude Discussion

Q. |
The probabilities that a student pass in Mathematics, Physics and Chemistry are $m$, $p$, $c$ respectively. Of these subjects the student has a 75 % chance of passing in at least one, a 50% chance of passing in at least two and 40 % chance of passing in exactly two. Which of the following relations are true ? |

✖ A. |
$p+m+c=\dfrac{19}{20}$ |

✔ B. |
$p+m+c=\dfrac{27}{20}$ |

✖ C. |
$pmc=\dfrac{1}{10}$ |

✖ D. |
$pmc=\dfrac{1}{4}$ |

**Solution:**

Option(**B**) is correct

Consider the figure shown below.

$p \cup m \cup c = 0.75$

$w + x + y = 0.4$

$z = 0.1$

$p \cup m \cup c = p + m + c - (w + x + y + 3z) + z$

$p + m + c = 1.35 =\dfrac{27}{20}$

**Edit:** Thank you **Arjun** for alternate explanation in the comments.

**Roop**

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**Arjun**

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*

Probability of non pass = 1 - Probability of at least one pass

$= 1 - 0.75 = 0.25$

$(1-m) (1-p) (1-c) = 0.25$

$(1 + mp - m -p) (1-c) = 0.25$

$1 + mp - m - p - c - mpc + mc + pc = 0.25$

$m + p + c - mp - pc - mc + mpc = 0.75$ -- (1)

Probability of exactly 2 pass $= 0.4$

$mp (1-c) + pc (1-m) + mc (1-p) = 0.4$

$mp + pc + mc - 3mpc = 0.4$

$mp + pc + mc - 2mpc = 0.5$ -- (2) (Adding the probability of all pass to probability of exactly 2 pass gives probability of at least 2 pass)

So, $mpc = 0.1$, -- (3)

From (2) and (3),

$mp + pc + mc - mpc = 0.6$ -- (4)

From (1) and (4),

$m + p + c = 0.75 + 06$

$m + p + c = 1.35$

$= \dfrac{135}{100}$

$= \dfrac{27}{20}$

Arjun's explanation is lucidly presented than the answer...😊