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A box contains exactly 22 balls, of which 11 are white and 11 are blac
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04 Dec 2023, 00:57
04 Dec 2023, 00:57
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A box contains exactly 22 balls, of which 11 are white and 11 are black. If two balls are to be picked from this box at random and without replacement, what is the probability that both balls will be black?
A. 1/6
B. 5/21
C. 5/11
D. 11/40
E. 17/40
A. 1/6
B. 5/21
C. 5/11
D. 11/40
E. 17/40
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Re: A box contains exactly 22 balls, of which 11 are white and 11 are blac
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04 Dec 2023, 20:46
04 Dec 2023, 20:46
1
There are 11 Black balls out 22 total balls. So, the probability of picking a Black the first time is 11/22. Now we have 10 Black balls out of 21 total balls. The probability of picking a Black is now 10/21.
\(\frac{11}{22} * \frac{10}{21}\)
\(= \frac{1}{2}*\frac{10}{21}\)
\(= \frac{5}{21}\)
\(\frac{11}{22} * \frac{10}{21}\)
\(= \frac{1}{2}*\frac{10}{21}\)
\(= \frac{5}{21}\)
A box contains exactly 22 balls, of which 11 are white and 11 are blac
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30 Jan 2024, 19:42
30 Jan 2024, 19:42
1
The probability of picking the black ball the first time is
\(\dfrac{11}{22}\)
The probability of picking the black ball the second times without replacement is
\(\dfrac{10}{21}\)
The probability that both balls will be black is
\(\dfrac{11}{22} \times \dfrac{10}{21} = \dfrac{1}{2} \times \dfrac{10}{21} = \dfrac{5}{21}\)
The answer is Choice B.
\(\dfrac{11}{22}\)
The probability of picking the black ball the second times without replacement is
\(\dfrac{10}{21}\)
The probability that both balls will be black is
\(\dfrac{11}{22} \times \dfrac{10}{21} = \dfrac{1}{2} \times \dfrac{10}{21} = \dfrac{5}{21}\)
The answer is Choice B.
