Discussion
P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches and lost 5% of the matches.
If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?
Answer: A
The probability is \(\frac{48}{125}\).
Each match is won with probability 0.8 = \(\frac{4}{5}\) and not won (drawn or lost) with probability 0.2 = \(\frac{1}{5}\). Exactly two wins in three matches is a binomial count:
\[ \begin{aligned} P &= \binom{3}{2}\left(\tfrac{4}{5}\right)^{2}\left(\tfrac{1}{5}\right) \\ &= 3\cdot\tfrac{16}{25}\cdot\tfrac{1}{5} = \tfrac{48}{125} \end{aligned} \]
\(\frac{16}{125}\) forgets the factor 3 for the three possible positions of the non-win, and \(\frac{16}{25}\) is just the chance of two particular wins. \(\frac{25}{48}\) is not a probability of this form.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development