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05 Mar 2019, 02:55
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
78% (01:13) correct
22%
(02:02)
wrong
based on 174
sessions
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A 6-sided die has 3 black sides and 3 white sides. If the die is thrown 4 times, what is the probability that, on at least one of the throws, the die will land with a black side up?
A. 1/16
B. 3/16
C. 1/2
D. 9/16
E. 15/16
A. 1/16
B. 3/16
C. 1/2
D. 9/16
E. 15/16
05 Mar 2019, 03:42
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A 6-sided die has 3 black sides and 3 white sides. If the die is thrown 4 times, what is the probability that, on at least one of the throws, the die will land with a black side up?
Solution:
Probability of at least one of the throws that the die will land with a black side up = 1- (Probability of having white side up all four times)
P(At least once black up) = 1- (3*3*3*3)/(6*6*6*6) => 1- (1/16) => 15/16
Answer must be 'E'
Solution:
Probability of at least one of the throws that the die will land with a black side up = 1- (Probability of having white side up all four times)
P(At least once black up) = 1- (3*3*3*3)/(6*6*6*6) => 1- (1/16) => 15/16
Answer must be 'E'
05 Mar 2019, 03:59
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Bunuel
Probability of a throw showing white at the top = 3/6 = 1/2
Probability of a throw showing Black at the top = 3/6 = 1/2
METHOD-1
Probability of die lying with black side up atleast once in 4 times = 1 - (probability of die to lie at white face up all 4 times)
i.e. Probability of die lying with black side up atleast once in 4 times \(= 1 - (\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}) = \frac{15}{16}\)
Answer: Option E
METHOD-2
Probability of 1 out of 4 throw to land on black side = \((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) * 4C1
Probability of 2 out of 4 throw to land on black side = \((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) * 4C2
Probability of 3 out of 4 throw to land on black side = \((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) * 4C3
Probability of 4 out of 4 throw to land on black side = \((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) * 4C4
i.e. Total Probability = (4C1 + 4C2 + 4C3 + 4C4) *\((\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2})\) = \(\frac{15}{16}\)
Answer: Option E
