Question

12th Probablity question.No spamming please. Brainliest answer rewarded.​

Answer 1

Step-by-step explanation:

Be

E1,E2 and E3 be the event that boxes I,II and III are chosen respectively.

The P(E1)=P(E2)=P(E3)=13

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

P(A/E1)=P(a gold coin from bag I)=22=1

P(A/E2)=P(a gold coin from bag II)=0

P(A/E3)=P(a gold coin from bag III)=12

By Baye's theorem

P(E1/A)=P(E1)P(A/E1)P(E1).P(A/E1)+P(E2)P(A/E2)+P(E3)P(A/E3)

⇒1/3×11/3×1+1/3×0+1/3×1/2

⇒2/3

Hence the required probability is 2/3

Answer 2

Answer:

Step-by-step explanation:

$\frac{1}{2}$ is the answer.Here, concept of Bay.es theorem is applied. In this theorem,we use rule of conditional probability.We select the main event and the subevents and the conditional probability is calculated. Selecting each box here has a one third probability. Similarly, picking probability of each silver coin from boxes is found.Finally the required probability is calculated using Baye.s theorem and we get half as answer.