A pinata breaks if it receives a strong hit or two medium hits. The probability of receiving a strong blow is 1/4, a medium blow 1/4 and the chance of a missed hit is 1/2. Four children are about to hit the pinata. What is the probability of it breaking?
Let P(S) be the probability of it breaking.Let P(D) be the possibility of a strong blow ,P(M) of a medium one and P(A) of missing.
I know I need to take into consideration the fact that I hit it twice, but my math seems to be failing me.
Edit: So, I thought that the possibility of missing four times is (1/4)^4 and the possibility of hitting it with medium power only once is 1/2^5.
Then, (1/2^4)+(1/2^5)=2/2^5. Now, P =1 – 2/2^5 = 15/16 =0.9375.
Second edit: If the chance of hitting with medium power is four times higher then it is 1/2^3. Which changes the result to 0.8125.
Best Answer
Hint: it will break unless it receive four misses or one medium hit. Compute the chance of each of these and subtract from $1$.
The Markov chain approach, which would be easier in some other problems, is to consider three states: original ($O$), damaged ($D$), broken ($B$). You have transition probabilities from each state to each other, which you can put in a matrix. The stating state is $\begin {pmatrix} 1\\0\\0 \end {pmatrix}$. Four multiplies by the matrix gets you the probability distribution after the four children.