5.8k views

The probabilities that a student passes in mathematics, physics and chemistry are $m,p$ and $c$ respectively. Of these subjects, the student has $75\%$ chance of passing in at least one, a $50\%$ chance of passing in at least two and a $40\%$ chance of passing in exactly two. Following relations are drawn in $m, p, c$:

1. $p + m + c = 27/20$
2. $p + m + c = 13/20$
3. $(p) \times (m) \times (c) = 1/10$
1. Only relation I is true.
2. Only relation II is true.
3. Relations II and III are true.
4. Relations I and III are true.

edited | 5.8k views

Probability of non pass $= 1 -$ Probability of at least one pass $= 1 - 0.75 = 0.25$

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

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

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

$\bf{m + p + c - mp - pc - mc + mpc = 0.75 \quad \to (1)}$

Probability of exactly $2$ pass $= 0.4$

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

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

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

So, $\bf{mpc = 0.1 \quad \to (3)}$

From $(2)$ and $(3),$
$\bf{mp + pc + mc - mpc = 0.6 \quad \to(4)}$

From $(1)$ and $(4)$

$m + p + c = 0.75 + 0.6$

$m + p + c = 1.35 = 135/100 = 27/20$

So, D option
by Veteran (431k points)
selected by
+2
@arjun sir, how did u come up with this solution ?

Also, how to approach such problems ?
0

@ravi_ssj4 You can also think it via Venn Diagram.

+1

option D is correct

P(M) = m, P (p) = p, P(c) = c
∵ The probability of at least one success
= P (M ∪ P ∪ C)
= m + p + c -mp - mc - pc + mcp = 3/4 ...(1)

The probability of at least two successes =
= mc(1 - p) + mp (1 - c ) + (1 - m )cp + mcp
= mc + mp + cp - 2mcp = 1/2

The probability of exactly two success
= mc(1 - p) + mp (1 - c ) +cp(1 - m )
= mc + mp + cp - 3 mcp = 2/5

(2) & (3) gives,

⇒ mcp = 1/2 - 2/5 = 1/10
∴ mc + mp + cp = 2/10 + 1/2 = 1/5 + 1/2 = 7/10

From (1),

m + p + c - 7/10 + 1/10 = 3/4
⇒ m + p + c = 3/4 + 7/10 - 1/10 = 27/20
Thus, pmc = 1/10 is a true relation.

by Active (3.3k points)
edited
0
Why have you written + "mcp" while calculating the probability of at least two successes
+3

atleast 2 means  >=2 i.e

= either 2 success or 3 success

= = mc(1 - p) + mp (1 - c ) + (1 - m )cp + mcp

0

Got it Thank you so much @Satbir

Sum of all M+P+C circles is given to be 0.75..(1)

Where each circle represents respective probability in passing.

The probability of passing in at least 2=0.5

The probability of passing in exactly 2=0.4

Therefore the probability of passing in exactly 3=0.5-0.4=0.1=mpc.

Now, Equation (1) can be restated as

$p+m+c-(mp+pc+mc)+mpc=0.75$............(2)

This is similar to

$n(A \cup B \cup C)=n(A)+n(B)+n(C)-(n(A \cap B)+n(B \cap C)+n(C \cap A))+n(A \cap B \cap C)$

Now to get blue area, which counts to 0.4, you would have taken $mp+pc+mc$ and to this subtracted the red part($mpc=0.1$), thrice,

hence to get mp+pc+mc, add $3 \times 0.1$ to 0.4

which gives

$mp+pc+mc=0.7$

plugging in values into (2) we get

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

And $mpc=0.1=p(m \cap p \cap c)=p \times m \times c$(A student passing or failing in subject is independent event).

So, III is true.

Ans- (D)

by Boss (29.1k points)

1