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A box contains 10 balls numbered from 1 to 10 inclusive. If
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19 Nov 2019, 11:23
19 Nov 2019, 11:23
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62% (00:37) correct
37% (00:46) wrong based on 83 sessions
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A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and then returns to the box, and then Jane removes a ball at random, what is the probability that both women removed the same ball?
A. \(\frac{1}{100}\)
B. \(\frac{1}{90}\)
C. \(\frac{1}{45}\)
D. \(\frac{1}{10}\)
E. \(\frac{41}{45}\)
A. \(\frac{1}{100}\)
B. \(\frac{1}{90}\)
C. \(\frac{1}{45}\)
D. \(\frac{1}{10}\)
E. \(\frac{41}{45}\)
Re: A box contains 10 balls numbered from 1 to 10 inclusive. If
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21 Nov 2019, 13:23
21 Nov 2019, 13:23
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For Ann or For Jane TOTAL cases = 10
For Ann favorable cases = 10
For Ann probability of picking the ball = 10/10 = 1
Now that she has replaced the ball TOTAL cases remains the same.
For Jane favorable case = 1 (If annhad picked the ball number 2, then jane has to pick only that ball, so only one choice for jane, hence only one fav case)
Probability for Jane picking the same ball = 1/10.
D is the right answer
For Ann favorable cases = 10
For Ann probability of picking the ball = 10/10 = 1
Now that she has replaced the ball TOTAL cases remains the same.
For Jane favorable case = 1 (If annhad picked the ball number 2, then jane has to pick only that ball, so only one choice for jane, hence only one fav case)
Probability for Jane picking the same ball = 1/10.
D is the right answer
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Re: A box contains 10 balls numbered from 1 to 10 inclusive. If
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08 Nov 2020, 09:15
08 Nov 2020, 09:15
1
huda wrote:
A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and then returns to the box, and then Jane removes a ball at random, what is the probability that both women removed the same ball?
A. \(\frac{1}{100}\)
B. \(\frac{1}{90}\)
C. \(\frac{1}{45}\)
D. \(\frac{1}{10}\)
E. \(\frac{41}{45}\)
A. \(\frac{1}{100}\)
B. \(\frac{1}{90}\)
C. \(\frac{1}{45}\)
D. \(\frac{1}{10}\)
E. \(\frac{41}{45}\)
P(Ann and Jane remove same ball) = P(Ann removes ANY ball AND Jane's ball matches Ann's ball)
= P(Ann removes ANY ball) x P(Jane's ball matches Ann's ball)
= 1 x 1/10
= 1/10
Answer: D
Aside: Once Jane removes her ball (and then replaces it), we have 10 balls, and 1 of them is the one that Jane picked.
So, P(Jane's ball matches Ann's ball) = 1/10
Cheers,
Brent
