What Is the Probability of Getting a Head on the First Toss If the Total Is Odd?

  • Thread starter Thread starter naspek
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary
SUMMARY

The probability of getting a "head" on the first toss of a fair coin, given that the total of three tosses is an odd number, can be calculated using conditional probability. The sample space for three tosses is represented as Ω = {0,1}^3, where 1 denotes heads and 0 denotes tails. The events E_1 (first toss is a head) and E_2 (total is odd) are defined, and the conditional probability P(E_1|E_2) can be computed using the formula P(E_1|E_2) = P(E_1 ∩ E_2) / P(E_2). This approach ensures a systematic calculation based on the defined events.

PREREQUISITES
  • Understanding of basic probability concepts
  • Familiarity with conditional probability
  • Knowledge of sample spaces in probability theory
  • Ability to compute intersections of events
NEXT STEPS
  • Study conditional probability in depth
  • Learn about sample space and event representation
  • Explore the concept of intersections in probability
  • Practice problems involving multiple coin tosses
USEFUL FOR

Students of probability theory, educators teaching statistics, and anyone interested in understanding the nuances of conditional probability in coin toss scenarios.

naspek
Messages
176
Reaction score
0
Hi all =)

Question...
A fair coin is tossed thrice. Supposed we denote a "head" turning up as 1 and
"tail" as 0. Given that the total on all three tosses is an odd number,
what is the probability that at first toss, we get a "head"?

i don't have any idea to start answering this question..
please guide me.. =)
 
Physics news on Phys.org


Ok our sample space is
[tex]\Omega = \{0,1\}^3 = \{(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)\}[/tex]
We assume a fair coin so every outcome is equally likely which gives us the probability function:
[tex]p(x) = \frac{1}{8} \qquad \textrm{for all }x \in \Omega[/tex]
Now we have two events [itex]E_1,E_2[/itex]. Let [itex]E_2[/itex] denote the event that the total on all three tosses is an odd number. Let [itex]E_1[/itex] denote the event that on the first toss we get a head. What you want is the conditional probability [itex]P(E_1|E_2)[/itex]. List all elements in [itex]E_1[/itex] and [itex]E_2[/itex] and you should be able to compute it using the formula:
[tex]P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}[/tex]
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 37 ·
2
Replies
37
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K