Question

A bag containing 4white 3 black and 2 yellow balls..a ball is drawn ,what is the probability that is white or black

Answer 1

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .Using Baye's theorem

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .Using Baye's theoremP(A∣C)=

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .Using Baye's theoremP(A∣C)= P(C∣A)∗P(A)+P(C∣B)∗P(B)

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .Using Baye's theoremP(A∣C)= P(C∣A)∗P(A)+P(C∣B)∗P(B)P(C∣A)∗P(A)

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .Using Baye's theoremP(A∣C)= P(C∣A)∗P(A)+P(C∣B)∗P(B)P(C∣A)∗P(A) = 1/2×(4/9)

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .Using Baye's theoremP(A∣C)= P(C∣A)∗P(A)+P(C∣B)∗P(B)P(C∣A)∗P(A) = 1/2×(4/9) 2 ×(5/9)

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .Using Baye's theoremP(A∣C)= P(C∣A)∗P(A)+P(C∣B)∗P(B)P(C∣A)∗P(A) = 1/2×(4/9) 2 ×(5/9) 3 +1/2×(1/3)

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .Using Baye's theoremP(A∣C)= P(C∣A)∗P(A)+P(C∣B)∗P(B)P(C∣A)∗P(A) = 1/2×(4/9) 2 ×(5/9) 3 +1/2×(1/3) 2 ×(2/3)

If A is the event of choosing the first bag and B the event of choosing the second bag. P(A)=P(B)=12Let C be the event of one white and 3 black being drawn .Using Baye's theoremP(A∣C)= P(C∣A)∗P(A)+P(C∣B)∗P(B)P(C∣A)∗P(A) = 1/2×(4/9) 2 ×(5/9) 3 +1/2×(1/3) 2 ×(2/3) 3 1/2×(4/9) 2 ×(5/9) 3 =125/287

Answer 2

Answer:

sorry I don't understand ur question your question is not at all clear can post your question clearly again