The probabilities that a student passes in Ma | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) Let $M,\,\,P$ and $C$ be the events of passing in mathematics, physics and chemistry respectively.

$P(M \cup P \cup C) = \frac{{75}}{{100}} = \

Vedclass · Accepted Answer

$p + m + c = \frac{{27}}{{20}}$

Vedclass · Answer

(b) Let $M,\,\,P$ and $C$ be the events of passing in mathematics, physics and chemistry respectively.

$P(M \cup P \cup C) = \frac{{75}}{{100}} = \frac{3}{4}$

$P(M \cap P) + P(P \cap C) + P(M \cap C) - 2P(M \cap P \cap C) = \frac{{50}}{{100}} = \frac{1}{2}$

$P(M \cap P) + P(P \cap C) + P(M \cap C) - 2P(M \cap P \cap C) = \frac{{40}}{{100}} = \frac{2}{5}$

$	herefore$ $m(1 - p)(1 - c) + p(1 - m)(1 - c) + c(1 - m)(1 - p)$

$ + mp(1 - c) + mc(1 - p) + pc(1 - m) + mpc = \frac{3}{4}$

$ \Rightarrow m + p + c - mc - mp - pc + mpc = \frac{3}{4}$ .....$(i)$

Similarly, $mp(1 - c) + pc(1 - m) + mc(1 - p) + mpc = \frac{1}{2}$

$ \Rightarrow mp + pc + mc - 2mpc = \frac{1}{2}$ .....$(ii)$

$mp(1 - c) + pc(1 - m) + mc(1 - p) = \frac{2}{5}$

$ \Rightarrow mp + pc + mc - 3mpc = \frac{2}{5}$ .....$(iii)$

From $(ii)$ to $(iii),$ $mpc = \frac{1}{2} - \frac{2}{5} = \frac{1}{{10}}$

From $(i)$ and $(ii),$ $m + p + c - mpc = \frac{3}{4} + \frac{1}{2}$

$	herefore \,\,\,m + p + c = \frac{3}{4} + \frac{1}{2} + \frac{1}{{10}} = \frac{{15 + 10 + 2}}{{20}} = \frac{{27}}{{20}}.$

Similar Questions

If $A$ and $B$ are any two events, then $P(\bar A \cap B) = $

If $A$ and $B$ are arbitrary events, then

Twelve tickets are numbered $1$ to $12$. One ticket is drawn at random, then the probability of the number to be divisible by $2$ or $3$, is

The probability of solving a question by three students are $\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{6}$ respectively. Probability of question is being solved will be

Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that One of them is black and other is red.