Samsung galaxy note 7 probability problem An unnamed smart-phone manufacturer has a problem with their phones bursting into flames.  The phones are shipped to stores in boxes of 12.  Before shipping,  a customer service agent tests 3 of the phones at random to ensure they won't suddenly combust.  If any of the three tested phones shows a sign of possibly burning, the shipment of twelve is held back.  If each phone has an independent probability of 20% that they show a sign of possibly burning, what percentage of shipments is held back?
 A: Probability that a phone is good is $0.8$. Probability that three tested phones are good is $0.8^3$ (assuming independence).
$P(\text{ship back})=1-P(\text{all 3 tested phones are good})=1-0.8^{3}= 48.8\%$
A: So, let $B$ denote "burned" and $N$ denote "non-burned".
We will have 4 possibilites:
$ \{ B, N, N \} $;
$  \{ B, B, N \} $;
$ \{ B,B,B \} $
And finally:
$ \{ N, N, N \} $
You know, from the problem, that $ P(B) = 0,2 $ and $P(A) = 0,8$. Doing one calculation, just as an example:
Probability of the 1st case:
$ P(1st) = 0,2 \cdot 0,8 \cdot 0,8 \cdot 6$
(The $6$ is in there because you have 6 possible ways of this case happen (changing the order of Ns and Bs)
I think this is enough for you to think about the problem and try it. Think about what the problem is asking you, and how can you calculate it.
A: There are four different possibilities:


*

*$P(X)$ - First phone fails the test

*$P(Y)$ - Second phone fails the test

*$P(Z)$ - Third phone fails the test

*$P(W)$ - No phone fails the test
As long as $P(W)$ doesn't come to happen, shipment is sent back. So question asks the answer to $1-P(W)$. Since probability of burning is $0.2$ probability of not burning is $0.8$.
$$P(W)=(\frac{8}{10})^{3}$$
$$1-(\frac{8}{10})^{3} = 0.488$$
