Samsung galaxy note 7 probability problem

Question

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?

Answer 1

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.

Answer 2

Probability that a phone is good is $0.8$. Probability that three tested phones are good is $0.8^3$ (assuming independence).

$P(\text{ship back})=1-P(\text{all 3 tested phones are good})=1-0.8^{3}= 48.8\%$

Answer 3

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 $0.8$.

$$P(W)=(\frac{8}{10})^{3}$$

$$1-(\frac{8}{10})^{3} = 0.488$$