shrive555 wrote:
A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball?
A. 1/100
B. 1/90
C. 1/45
D. 1/10
E. 41/45
P(Ann selecting a ball) = \(\frac{1}{10}\)
P(jane selecting a ball) =\(\frac{1}{10}\)
P(both removed the same ball) = \(\frac{1}{10}\) *\(\frac{1}{10}\)= \(\frac{1}{100}\)
isn't correct ?
METHOD-1No. of ways for Ann to select a ball out of 10 balls = 10C1 = 10 ways
No. of ways for Jane to select Same ball that ann select = 1C1 = 1 ways
Total Favourable ways of Ann and Jane to pick the same ball = 10*1 = 10 ways
Total ways of Ann and Jane to pick balls randomly = 10*10 = 100 ways
Probability = Favourable Outcomes / Total Outcomes = 10/100 = 1/10
Answer: option D
METHOD-2Probability of Ann to select any ball out of 10 balls = 10/10 = 1
Probability of Jane to select Same ball that Ann selected out of a total 10 balls = 1/10
Total Probability of Both events to happen = 1*(1/10) = 1/10
Answer: option D
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57% (00:43) correct 
