Originally Posted by

**Sudharaka** Dear entropyyy,

Your idea is correct but your expression is incorrect. You have to **add** the probability of all three turning left.

That is, Probability that at least two cars turning left= $\displaystyle (0.7 \times 0.7 \times 0.3) + (0.7)^3$

Although the answer is obtained I would like you to show how the problem could be solved in a more methodical way.

Using our abbrevations,

Event that at least two cars turn left= $\displaystyle (A\cap{B}\cap{C})\cup{(A\cap{B}\cap{D}})}$

Since the two events $\displaystyle A\cap{B}\cap{C}~and~A\cap{B}\cap{D}$ are mutually exclusive,

Probability that at least two cars turn left= $\displaystyle P(A\cap{B}\cap{C}) + P(A\cap{B}\cap{D})$

Since A,B,C and A,B,D are independent,

Probability that at least two cars turn left= $\displaystyle P(A)\timesP(B)\timesP(C) + P(A)\timesP(B)\times(D)$

Probability that at least two cars turn left= $\displaystyle (0.7\times0.7\times0.7)+(0.7\times0.7\times0.3)$

Hope this will help you.