GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
The probability is that a certain coin will turn up heads on
[#permalink]
30 Jul 2018, 10:42
30 Jul 2018, 10:42
1
Expert Reply
00:00
Question Stats:
92% (00:27) correct
7% (00:50) wrong based on 92 sessions
Hide Show timer Statistics
The probability is \(\frac{1}{2}\) that a certain coin will turn up heads on any given toss and the probability is \(\frac{1}{6}\) that a number cube with faces numbered 1 to 6 will turn up any particular number. What is the probability of turning up a heads and a 6?
A. \(\frac{1}{36}\)
B. \(\frac{1}{12}\)
C. \(\frac{1}{6}\)
D. \(\frac{1}{4}\)
E. \(\frac{2}{3}\)
A. \(\frac{1}{36}\)
B. \(\frac{1}{12}\)
C. \(\frac{1}{6}\)
D. \(\frac{1}{4}\)
E. \(\frac{2}{3}\)
Re: The probability is that a certain coin will turn up heads on
[#permalink]
07 Aug 2018, 06:04
07 Aug 2018, 06:04
2
Expert Reply
Explanation
The probability of independent events A and B occurring is equal to the product of the probability of event A and the probability of event B.
In this case, the probability of the coin turning up heads is \(\frac{1}{2}\) and the probability of rolling a 6 is \(\frac{1}{6}\). Therefore, the probability of heads and a 6 is equal to \(\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}\).
Alternatively, list all the possible outcomes: H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6.
There are 12 total outcomes and only 1 with heads and a 6. Therefore, the desired outcome divided by the total number of outcomes is equal to \(\frac{1}{12}\).
The probability of independent events A and B occurring is equal to the product of the probability of event A and the probability of event B.
In this case, the probability of the coin turning up heads is \(\frac{1}{2}\) and the probability of rolling a 6 is \(\frac{1}{6}\). Therefore, the probability of heads and a 6 is equal to \(\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}\).
Alternatively, list all the possible outcomes: H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6.
There are 12 total outcomes and only 1 with heads and a 6. Therefore, the desired outcome divided by the total number of outcomes is equal to \(\frac{1}{12}\).
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1115
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Re: The probability is that a certain coin will turn up heads on
[#permalink]
14 Oct 2022, 09:47
14 Oct 2022, 09:47
1
We are given that a fair coin is tossed and a fair dice is rolled and We need to find What is the probability of turning up a heads and a 6?
Probability (Getting a Head and a 6) = P(H) * P(6) = \(\frac{1}{2}\) * \(\frac{1}{6}\) (As there is 1 out of 6 ways in which we can get a 6 out of 1 to 6)
= \(\frac{1}{12}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
Watch the following video to learn How to Solve Dice Rolling Probability Problems
Probability (Getting a Head and a 6) = P(H) * P(6) = \(\frac{1}{2}\) * \(\frac{1}{6}\) (As there is 1 out of 6 ways in which we can get a 6 out of 1 to 6)
= \(\frac{1}{12}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
Watch the following video to learn How to Solve Dice Rolling Probability Problems
