Events & Promotions
|
|
GMAT 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.
Jul 1011:00 AM PDT
-11:00 AM PDT
In our conversation, Kishore talks about how strategically he used data analytics to identify high impact areas to focus on as well as his early morning study routine (3 AM prep!) that optimized his GMAT preparation.- Jun 30
08:00 AM PDT
-11:59 PM PDT
Join us June 30th for 2 weeks of GMAT questions (just 8 per day) to help with your prep and to compete with your fellow GMAT Clubbers for a chance to win a prize fund of $30,000 for you and your team-mates! 
Jul 0901:00 PM EDT
-02:00 PM EDT
More Target Test Prep students are earning 100th percentile GMAT scores. Don’t settle for less when the Target Test Prep course gives you everything you need to score high on the GMAT. Start your 5-day free trial today.
Jul 1206:30 AM PDT
-08:30 AM PDT
Join our webinar to master GMAT RC Main Point Questions! Learn effective strategies to decipher passage themes and purposes. Senior Verbal Expert- Sunita Singhvi will equip you with tools to navigate complexities and avoid common traps set by the GMAT.
Jul 1211:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
Jul 1311:00 AM EDT
-12:30 PM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Jul 1508:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Jul 1611:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
14 Feb 2020, 11:08
Kudos
Bookmarks
marcodonzelli
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?
A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32
A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32
The probability that the first 3 flips are heads and last 2 flips are tails is:
P(HHHTT) = (1/2)^5 = 1/32.
Answer: E
13 Apr 2022, 04:27
Kudos
Bookmarks
Bunuel
marcodonzelli
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?
A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32
A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32
We want the probability that coin will land heads up on the first 3 flips and not on the last 2 flips. So there is ONLY one favorable outcome, namely heads up on the first 3 flips and tails up on the last 2 flips: HHHTT.
# of total out comes is 2^5=32.
P=favorable/total=1/32.
Answer: E.
IF THE QUESTION WERE:
A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)
Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\).
As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question.
Hope it's clear.
Hi brunel, great info always. Thought the repitition is taken care of by 3!2! already. Therefore not sure why are we multiplying (1/2)^3∗(1/2)^2 again? Am I missing something? Could you explain? Thanks
P=5!/3!2!∗(1/2)^3∗(1/2)^2
13 Apr 2022, 18:08
Kudos
Bookmarks
Kimberly77
Bunuel
marcodonzelli
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?
A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32
A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32
We want the probability that coin will land heads up on the first 3 flips and not on the last 2 flips. So there is ONLY one favorable outcome, namely heads up on the first 3 flips and tails up on the last 2 flips: HHHTT.
# of total out comes is 2^5=32.
P=favorable/total=1/32.
Answer: E.
IF THE QUESTION WERE:
A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)
Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\).
As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question.
Hope it's clear.
Hi brunel, great info always. Thought the repitition is taken care of by 3!2! already. Therefore not sure why are we multiplying (1/2)^3∗(1/2)^2 again? Am I missing something? Could you explain? Thanks
P=5!/3!2!∗(1/2)^3∗(1/2)^2
The (1/2)^3*(1/2)^2 is the probability of 3 heads and 2 tails occurring in one specified pattern or way, in 5 tosses.
5!/3!2! is the number of ways to create 3 heads and 2 tails in 5 tosses.
5! is the number of arrangements of 5 different things.
Since 3 heads are being treated above as if they were 3 different things with 3! arrangements, where in fact they are just one arrangement, the 5! needs to be divided by 3!. The same logic applies to the 2 tails.
Posted from my mobile device
