Probability problem: $4$ kids hitting a balloon

Question

Let's say there is a balloon that blows if it gets a hard hit or 2 medium hits. During a party, if a kid hitsthe balloon, he has $\frac{1}{4}$ probability getting a hard hit on the balloon and $\frac{1}{4}$ probability for a medium hit and $\frac{1}{2}$ probability of missing the hit. If $4$ kids consecutively hit the balloon(a kid can hit only one time), what's the probability of the balloon to blow?

All hits are independent.

Any hint would be valuable.

Answer 1

There are two ways to fail: all kids hit empty or 1 medium. The probability of the former is $\frac{1}{16}$, the latter is $\frac{1}{8}$. So what you need is $1-\frac{1}{8}-\frac{1}{16}=\frac{13}{16}$

Answer 2

Try consider: number of hard hits cannot exceed one, and the number of medium hits cannot exceed $2$. You can solve this question by dividing it into cases. Another hint: consider the probability of balloon not blowing? You then have to rule out hard hits or $>1$ medium hits.

Answer 3

I will denote a miss as $-$

The possible scenarios in which the ballon will blow up are as follows:

We can get $1$ hard in ${4 \choose 1}=4$ different ways

$${H},{-H},{--H},{---H}$$

We can get $2$ mediums in ${4 \choose 2}=6$ different ways

$$MM,M-M,M--M,-MM,-M-M,--MM$$

We can get $1$ hard and $1$ medium (with medium preceding hard) in ${4 \choose 2}=6$ different ways

$$MH,M-H,M--H,-MH,-M-H,--MH$$

One can use multiplicity to find each of these probabilities.

Summing all these probabilities, you will find that $$p=\frac{13}{16}$$

which agrees with Alex's $[+1]$ more elegant solution.

Answer 4

Introduce four i.i.d. random variables $X_i, i=1\dots4$, with $p(X_i=0)={1\over 2}$, $p(X_i={1\over 2})={1\over 4}$, $p(X_i=1)={1\over 4}$. $X_i=0$ represents a miss, $X_i={1\over 2}$ is a medium hit, $X_i=1$ is a hard hit.

With the assumption that damage accumulates (hits by different kids do blow up the balloon), introduce the sums $S_n=\sum_{i=1}^n X_i, n=1\dots 4$. $S_n$ is the amount of damage the balloon suffered after $n$ kids had a go at it, and the balloon blows up when $S_n$ crosses $1$. Now, all you need to do is to calculate $1-p(S_4<1)$, i.e. one minus the probability that all four kids don't blow up the balloon. Since the $X_i$ are non-negative this is the same as $p(S_4\geq 1)$ (translated: it makes no difference if the kids keep hitting the balloon even though it already blew up).

Is this enough of a hint?