Question

An expert burglar has 8 master keys to open several new houses of a mansion.Only one master key will open a given house.If $40\%$ of these houses remain unlocked the probability that the burglar will break open the treasury house on selecting 3 master keys at random is?

• A. 5/8
• B. 2/8
• C. 3/5
• D. 3/4

Answer 1

The correct option is A

5/8

X: Home is unlocked=P(X)=$\frac{2}{5}$

Y:Home is locked:P(Y)=$\frac{3}{5}$

Z: Room gets opened by any of the 3 keys

P(Z)=P(Z intersection X)+P(Z intersection Y)

$P\left(Z\right)=P\left(X\right)\times P\left(\frac{Z}{X}\right)+P\left(Y\right)\times P\left(\frac{Z}{Y}\right)$

$P\left(Z\right)=\frac{2}{5}+\frac{3}{5}\times P\left(\frac{Z}{Y}\right)$

$P\left(\frac{Z}{Y}\right)=\frac{{}^7{C}_2}{{}^8{C}_3}=\frac{3}{8}$

Hence $P\left(Z\right)=\frac{2}{5}+(\frac{3}{5}\times \frac{3}{8})=\frac{5}{8}$