A carton contains 12 green and 8 blue bulbs. 2 bulbs are drawn at random. Find the probability that they are of same colour?A. (91/47)B. (47/105)C. (47/95)D. (47/145) answer with explanation About the author Parker
Answer: Let S be the sample space Then n(S) = no of ways of drawing 2 bulbs out of (12+8) = 20c2=20*19/2*1=190 Let E = event of getting both bulbs of same colour Then, n(E) = no of ways (2 bulbs out of 12) or (2 bulbs out of 8) =12C2+ 8C2=(132/2)+(56/2) = 66+28 = 94 Therefore, P(E) = n(E)/n(S) = 94/190 = 47/95 Step-by-step explanation: Reply
Step-by-step explanation: Let S be the sample space Then n(S) = no of ways of drawing 2 bulbs out of (12+8) = 20c2=20*19/2*1=190 Let E = event of getting both bulbs of same colour Then, n(E) = no of ways (2 bulbs out of 12) or (2 bulbs out of 8) =12C2+ 8C2=(132/2)+(56/2) = 66+28 = 94 Therefore, P(E) = n(E)/n(S) = 94/190 = 47/95 Reply
Answer:
Let S be the sample space
Then n(S) = no of ways of drawing 2 bulbs out of (12+8) = 20c2=20*19/2*1=190
Let E = event of getting both bulbs of same colour
Then, n(E) = no of ways (2 bulbs out of 12) or (2 bulbs out of 8)
=12C2+ 8C2=(132/2)+(56/2) = 66+28 = 94
Therefore, P(E) = n(E)/n(S) = 94/190 = 47/95
Step-by-step explanation:
Step-by-step explanation:
Let S be the sample space
Then n(S) = no of ways of drawing 2 bulbs out of (12+8) = 20c2=20*19/2*1=190
Let E = event of getting both bulbs of same colour
Then, n(E) = no of ways (2 bulbs out of 12) or (2 bulbs out of 8)
=12C2+ 8C2=(132/2)+(56/2) = 66+28 = 94
Therefore, P(E) = n(E)/n(S) = 94/190 = 47/95