Question

Q 17 Ryan has 10 gold coins in his one pocket
(say A) and 5 silver and 5 gold coins in
the other pocket (say B). He randomly
chooses a pocket and picks a coin from
it. If the coin he chose was gold, then
what is the probability that he chose the
coin from pocket A?
ps: A
3/4
В.
1/2​

Answer 1

P(A)=10/10=1
P(B)=5/10=1/2

P(G)=1/2*1+1/2*1/2=3/4
P(G/A)=(1/2*1)/3/4=2/3
Answer is 2/3

Answer 2

Correct Question

Ryan has 10 gold coins in his one pocket (say A) and 5 silver and 5 gold coins in the other pocket (say B). He randomly chooses a pocket and picks a coin from it. If the coin he chose was gold, then what is the probability that he chose the coin from pocket A?

A. $\frac{3}{4}$

B. $\frac{1}{2}$

C. $\frac{2}{3}$

Answer:

The probability that he chose the coin from pocket A is $\frac{2}{3}$.

Step-by-step explanation:

Given:

Ryan has 10 gold coins in his one pocket (say A).

5 silver and 5 gold coins in the other pocket (say B).

If the coin he chose was gold.

To Find:

The probability that he chose the coin from pocket A.

Solution:

Let $E_{A$ and $E_{B}$ are the events that the Pocket A and  B are chosen respectively.

Probability of $E_{A$,  $P(E_{A} ) =\frac{1}{2}$

Probability of $E_{B$,  $P(E_{B} ) =\frac{1}{2}$

Let K be the event that the coin drawn is of gold.

Probability of a gold coin from Pocket A$=P(K}|{E_{A} )$

 $P(K}|{E_{A} )=\frac{10}{10} =1$

Probability of a gold coin from Pocket B$=P(K}|{E_{B} )$

$P(K}|{E_{B} )=\frac{5}{10} =\frac{1}{2}$

The probability that the gold coin is drawn from Pocket A $=P({E_{A} |K )$

By using Bayes' theorem,

$P(E_{A}|K ) =\frac{P(E_{A})\times P(K |E_{A} )}{P(E_{A})\times P(K |E_{A} )+P(E_{B})\times P(K |E_{B} )}$

               $=\frac{\frac{1}{2} \times1}{\frac{1}{2} \times1+\frac{1}{2} \times \frac{1}{2} }$

               $=\frac{\frac{1}{2} }{\frac{1}{2} +\frac{1}{4} }$    

                 $=\frac{\frac{1}{2} }{\frac{3}{4} }=\frac{2}{3}$

$P({E_{A} |K )=\frac{2}{3}$

Thus,the probability that he chose the coin from pocket A is $\frac{2}{3}$.

#SPJ3