If a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exectly twice in 3 consecutive flips ?
A. 0.125
B. 0.25
C. 0.375
D. 0.5
E. 0.666
OA C
Source: GMAT Prep
If a coin has an equal probability of landing heads up or ta
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P(H) = 1/2.BTGmoderatorDC wrote:If a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exectly twice in 3 consecutive flips ?
A. 0.125
B. 0.25
C. 0.375
D. 0.5
E. 0.666
P(T) = 1/2.
One way to get exactly 1 heads in 3 flips:
P(HTT) = 1/2 * 1/2 * 1/2 = 1/8.
Since H can appear in a total of 3 positions -- 1st flip, 2nd flip, or 3rd flip -- we multiply by 3:
1/8 * 3 = 3/8 = 0.375.
The correct answer is C.
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We need to determine the probability of HHT and its variations:BTGmoderatorDC wrote:If a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exectly twice in 3 consecutive flips ?
A. 0.125
B. 0.25
C. 0.375
D. 0.5
E. 0.666
OA C
Source: GMAT Prep
P(HHT) = (1/2)^3 = 1/8
Since we can arrange HHT in 3!/2! = 3 ways, the overall probability is 3 x 1/8 = 3/8 = 0.375.
Answer: C
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The probability of coin landing heads up twice in three consecutive flips will have three possibilities.
$$HHT,\ HTH\ or\ THH$$
$$HHT=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$$
$$HTH=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$$
$$THH=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$$
Add up the three
$$\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}=0.375$$
$$answer\ is\ Option\ C$$
$$HHT,\ HTH\ or\ THH$$
$$HHT=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$$
$$HTH=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$$
$$THH=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$$
Add up the three
$$\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}=0.375$$
$$answer\ is\ Option\ C$$