Probability Ques 63
The probabilities that a student passes in Mathematics, Physics and Chemistry are $m, p$ and $c$, respectively. Of these subjects, the students has a $75 \%$ chance of passing in atleast one, a $50 \%$ chance of passing in atleast two and a $40 \%$ chance of passing in exactly two. Which of the following relations are true?
(a) $p+m+c=\frac{19}{20}$
(b) $p+m+c=\frac{27}{20}$
(c) $p m c=\frac{1}{10}$
(d) $p m c=\frac{1}{4}$
$(1999,3 M)(2011)$
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Answer:
Correct Answer: 63.(b,c)
Solution:
Formula:
Set theoretical notation of probability and some important results:
- Let $A, B$ and $C$ respectively denote the events that the student passes in Maths, Physics and Chemistry.
It is given,
$P(A)=m, P(B)=p$ and $P(C)=c$ and
$P$ (passing atleast in one subject)
$ =P(A \cup B \cup C)=0.75 $
$\Rightarrow \quad 1-P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right)=0.75$
$ \because \quad[P(A)=1-P(\bar{A}) $
and $\quad\left[P(\overline{A \cup B \cup C}]=P\left(A^{\prime} \cap B \cap C^{\prime}\right)\right]$
$\Rightarrow 1-P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right) \cdot P\left(C^{\prime}\right)=0.75$
$\because A, B$ and $C$ are independent events, therefore $A^{\prime}, B$ and $C^{\prime}$ are independent events.
$\Rightarrow 0.75 =1-(1-m)(1-p)(1-c) $
$\Rightarrow 0.25 =(1-m)(1-p)(1-c)$ $\quad …….(i)$
Also, $P$ (passing exactly in two subjects) $=0.4$
$\Rightarrow P(A \cap B \cap \bar{C} \cup A \cap \bar{B} \cap C \cup \bar{A} \cap B \cap C)=0.4$
$\Rightarrow P(A \cap B \cap \bar{C})+P(A \cap \bar{B} \cap C)+P(\bar{A} \cap B \cap C)=0.4$
$\Rightarrow P(A) P(B) P(\bar{C})+P(A) P(\bar{B}) P(C)$
$ +P(\bar{A}) P(B) P(C)=0.4 $
$\Rightarrow \quad p m(1-c)+p(1-m) c+(1-p) m c=0.4$
$\Rightarrow \quad p m-p m c+p c-p m c+m c-p m c=0.4$ $\quad$ …….(ii)
Again, $P$ (passing atleast in two subjects) $=0.5$
$\Rightarrow P(A \cap B \cap \bar{C})+P(A \cap \bar{B} \cap C)$
$+P(\bar{A} \cap B \cap C)+P(A \cap B \cap C)=0.5$
$\Rightarrow \quad p m(1-c)+p c(1-m)+c m(1-p)+p c m=0.5$
$\Rightarrow p m-p c m+p c-p c m+c m-p c m+p c m=0.5$
$\Rightarrow \quad(p m+p c+m c)-2 p c m=0.5\quad \quad $ $\quad$ …….(iii)
From Eq. (ii),
$\Rightarrow \quad p m+p c+m c-3 p c m=0.4 \quad\quad$ $\quad$ …….(iv)
From Eq. (i),
$0.25= 1-(m+p+c)+(pm+pc+cm)-pcm$ $\quad$ …….(v)
On solving Eqs. (iii), (iv) and (v) we get
$p + m+ c = 1.35 = 27/20$
Therefore, option (b) is correct.
Also, from Eqs. (ii) and (iii), we get pmc=$1/10$
Hence, option (c) is correct
BETA
