Last visit was: 14 Jul 2025, 02:26 It is currently 14 Jul 2025, 02:26
Close
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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 13 Jul 2025
Posts: 102,569
Own Kudos:
741,196
 [5]
Given Kudos: 98,178
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,569
Kudos: 741,196
 [5]
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
User avatar
Dereno
Joined: 22 May 2020
Last visit: 14 Jul 2025
Posts: 451
Own Kudos:
415
 [1]
Given Kudos: 364
Products:
Posts: 451
Kudos: 415
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
ManifestDreamMBA
Joined: 17 Sep 2024
Last visit: 13 Jul 2025
Posts: 977
Own Kudos:
634
 [1]
Given Kudos: 192
Products:
Posts: 977
Kudos: 634
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
GraemeGmatPanda
Joined: 13 Apr 2020
Last visit: 11 Jul 2025
Posts: 70
Own Kudos:
Given Kudos: 18
Location: United Kingdom
Concentration: Marketing
Products:
Posts: 70
Kudos: 110
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Identify the event where the first coin shows heads and the second coin shows tails and use independence to find its probability ('AND' rule)
\(P(H_{1}\text{ and }T_{2})=0.6\times0.4\)

Identify the event where the first coin shows tails and the second coin shows heads and use independence to find its probability. ('AND' rule)
\(P(T_{1}\text{ and }H_{2})=0.4\times0.6\)

Since the two scenarios are mutually exclusive, add their probabilities to obtain the total probability of one head and one tail ('OR' rule).
\(P(\text{one head and one tail})=0.6\times0.4+0.4\times0.6=0.24+0.24=0.48\)

Answer E
ʕ•ᴥ•ʔ
Moderators:
Math Expert
102569 posts
PS Forum Moderator
691 posts