Question

A box contains 15 resistors. twelve of them are labelled 50ω and the other three are labeled 100ω. what is the probability that the second resistor is 100ω, given that the first resistor is 50ω?

Answer 1

Answer : P(second resistor is 100ω , given that the first resistor is 50ω) is given by

$\frac{1}{5}$

Explanation :

Since we have given that

Total number of resistors =15

Number of resistors labelled with 50ω = 12

Number of resistors labelled with 100ω =3

Let A: Event getting resistor with 50ω

B: Event getting resistor with 100ω

Since A and B are independent events .

So,

$P(A\cap B)=P(A).P(B)$

Now, According to question , we can get that

$P(A)= \frac{12}{15}=\frac{4}{5}\\\\P(B)=\frac{3}{15}=\frac{1}{5}$

So,

$P(A\cap B)=P(A).P(B)\\\\P(A\cap B)=\frac{4}{5}\times \frac{1}{5}\\\\P(A\cap B)=\frac{4}{25}$

So, by using the conditional probability , which state that

$P(B\mid A)=\frac{P(A\cap B)}{P(A)}$

$P(B\mid A)=\frac{\frac{4}{25}}{\frac{4}{5}}\\\\P(B\mid A)=\frac{5}{25}\\\\P(B\mid A)=\frac{1}{5}$

So, P(second resistor is 100ω , given that the first resistor is 50ω) is given by

$\frac{1}{5}$