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John flips a quarter coin twice
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12 Jan 2021, 09:17
12 Jan 2021, 09:17
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Question Stats:
65% (00:27) correct
34% (00:30) wrong based on 38 sessions
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John flips a quarter coin twice
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
The chances of John getting two tails |
The chances of John getting NO tails |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
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Posts: 6218
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Re: John flips a quarter coin twice
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12 Jan 2021, 10:31
12 Jan 2021, 10:31
1
Carcass wrote:
John flips a quarter coin twice
Quantity A |
Quantity B |
The chances of John getting two tails |
The chances of John getting NO tails |
The correct answer here is actually D, since we're not told that this is a "fair" coin (i.e., on any toss, the probability of heads = probability of tails = 0.5)
IF we assume that P(heads) = P(tails) = 0.5, then...
P(two tails) = P(tails on first toss AND tails on second toss)
= P(tails on first toss) x) P(tails on second toss)
= 0.5 x 0.5
= 0.25
P(no tails) = P(2 heads)
P(heads on first toss AND heads on second toss)
= P(heads on first toss) x) P(heads on second toss)
= 0.5 x 0.5
= 0.25
Answer: C
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John flips a quarter coin twice
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23 Oct 2022, 09:10
23 Oct 2022, 09:10
Given that John flips a quarter coin twice
Quantity A: The chances of John getting two tails
As we are flipping a coin twice so there are \(2^2\) = 4 cases
HH, HT, TH, TT
The chances of John getting two tails = \(\frac{1}{4}\)
(As there is only one case TT out of 4 in which John can get two Tails)
Quantity B: The chances of John getting NO tails
The chances of John getting NO Tails = Chance of getting 2 heads= \(\frac{1}{4}\)
(As there is only one case HH out of 4 in which John can get NO tails or 2 Heads)
Clearly, Quantity A = Quantity B = \(\frac{1}{4}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
Quantity A: The chances of John getting two tails
As we are flipping a coin twice so there are \(2^2\) = 4 cases
HH, HT, TH, TT
The chances of John getting two tails = \(\frac{1}{4}\)
(As there is only one case TT out of 4 in which John can get two Tails)
Quantity B: The chances of John getting NO tails
The chances of John getting NO Tails = Chance of getting 2 heads= \(\frac{1}{4}\)
(As there is only one case HH out of 4 in which John can get NO tails or 2 Heads)
Clearly, Quantity A = Quantity B = \(\frac{1}{4}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
