Maths Answer the following questions in short.

∴ The probability of Y solving the problem is

P(Y) = 1 – P(Y’)

$\begin{aligned} & =1-\frac{2}{2+3} \\ & =1-\frac{2}{5} \\ & =\frac{3}{5}\end{aligned}$

∴ The probability of X solving the problem is

$P(X)=\frac{4}{4+3}=\frac{4}{7}$

$\therefore P ( A )=\frac{{ }^7 C _1}{{ }^{12} C _1}=\frac{7}{12}$

Let event B: The second pencil chosen is yellow.

Since the first yellow pencil is not replaced in the box, we now have 11 pencils, out of which 6 are yellow.

∴ Probability that the second pencil is yellow under the condition that the first yellow pencil is not replaced in the box = P(B/A)

$\begin{aligned} & =\frac{{ }^6 C_1}{{ }^{11} C_1} \\ & =\frac{6}{11}\end{aligned}$

Required probability

= P(A ∩ B) = P(B/A) . P(A)

$\begin{aligned} & =\frac{6}{11} \times \frac{7}{12} \\ & =\frac{7}{22}\end{aligned}$

$\therefore P(A)=\frac{{ }^3 C _1}{{ }^7 C _1}=\frac{3}{7}$

Let event B: The second marble drawn is blue.

Since the first red marble is not replaced in the bag, we now have 6 marbles out of which 4 are blue.

∴ Probability that the second marble is blue under the condition that the first red marble

is not replaced in the bag $= P ( B / A )=\frac{{ }^4 C _1}{{ }^6 C _1}=\frac{4}{6}=\frac{2}{3}$

∴ Required probability = P(A ∩ B) = P(B/A) . P(A)

$\begin{aligned} & =\frac{2}{3} \times \frac{3}{7} \\ & =\frac{2}{7}\end{aligned}$

Alternate Method:

Total number of marbles = 3 + 4 = 7

Two marbles are drawn at random without replacement.

$\therefore n ( S )={ }^7 C _1 \times{ }^6 C _1=7 \times 6=42$

Let event A: The first marble is red and second marble is blue.

First red marble can be drawn from 3 red marbles in ${ }^3 C_1$ ways and second blue marble

can be drawn from 4 blue marbles in ${ }^4 C_1$ ways.

$\begin{aligned} & \therefore n ( A )={ }^3 C _1 \times{ }^4 C _1=3 \times 4=12 \\ & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{12}{42}=\frac{2}{7}\end{aligned}$