Question

A real estate man has eight master keys to open several new houses. Only one master key will open a given house. If 40% of these homes are usually left unlocked, the probability that the real estate man can get into a specific home if he selects three master keys at random, is

• A. 1/2
• B. 5/8
• C. 2/3
• D. 3/4

Answer 1

Answer: (B)

U : Home is unlocked: $P(U)=2/5$
L : Home is locked; $P(L)=3/5$
A : room is opened by any of the 3 keys

$P(A)=P(A\cap U)+P(A\cap L)=P(U)\cdot P(A/U)+P(L)\cdot P(A/L)$
$P(A)=\frac{2}{5}\cdot 1+\frac{3}{5}\cdot P(A/L)$
$P(A/L)=\frac{{}^7{C}_2}{{}^8{C}_3}=\frac{21}{56}=\frac{3}{8}\left[k_1{k}_2{k}_3{k}_4{k}_5{k}_6{k}_7{k}_8\right]$
$\therefore P(A)=\frac{2}{5}+\frac{3}{5}\cdot \frac{3}{8}=\frac{2}{5}+\frac{9}{40}=\frac{16+9}{40}=\frac{5}{8}$