Question

New codes to destroy the First Order are known to exist. With probability 0.6, the codes are hidden by Leia; with probability 0.4, they are hidden by Han Solo. When Leia hides the codes, she hides them with BB-8 70% of the time and with C-3PO 30% of the time. Han Solo is equally likely to hide them with BB-8 and C-3PO a. What is the probability that the codes are with BB-8? b. Given that the codes are with C-3PO, what is the probability the codes were hidden by Han Solo?

Answer 1

Answer: (a). 0.62 (b). 0.53

Step-by-step explanation:

Let us start by defining some events seen:

Given;

L rep the event that codes hidden Leia

BB represents the codes hidden in BB-8

H is the event that codes hidden by Hah

C-3PO rep the codes hidden in C3PO

Giving the probabilities, we have that

P(L)  =  probability that codes are hidden by Leia = 0.6

P(H) = probability that codes are hidden by Hah = 0.4

where P(BB/L) = 70/10 i.e 70%

P(C-3PO/L) = 1 - P(BB/L) = 1 - 7/10 = 3/10

Also, P(BB/H)  = 1/2 = P(C-3PO/H)

We can say that Han Solo is likely to hide them in aceta BB-8 and C-3PO

To begin,

(a). The question tell us to find the probability that the codes are with BB-8.

First of all, we find P(BB).

P(BB) = P(L)*P(BB/L) + P(H)*P(BB/H)

solving this we get

P(BB) = 0.6*7/10 + 0.4*1/2 = 0.62

Therefore,  the probability that the codes are with BB-8 = 0.62

(b). The second question says;

Given that the codes are with C-3PO, what is the probability the codes were hidden by Han Solo?

This tell us to find P(H/C-3PO)

P(H/C-3PO) = P(C-3PO/H)*P(H) / P(C-3PO/H) + P(C-3PO/L)*P(L)

inputting the values gives us;

P(H/C-3PO) = 1/2 * 0.4 / ( 1/2 *0.4 + 3/10 *0.6) = 0.53

SO we have that the probability the codes were hidden by Han Solo = 0.53

cheers I hope this was helpful!!