Understanding the Chess Match Probability Problem
This question asks for the probability of a specific outcome in a series of independent events. We are given the probabilities of P winning, drawing, and losing against Q in a single chess match. We need to find the probability that P wins exactly 2 out of the next 3 matches played.
Defining Probabilities and Outcomes
Let's define the probabilities for a single match:
- Probability of P winning (let's call this 'p'): $p = 80\% = 0.80$
- Probability of a draw: $15\% = 0.15$
- Probability of P losing: $5\% = 0.05$
For this problem, we are interested in whether P wins or does not win. The event "P does not win" includes draws and losses.
Probability of P not winning (let's call this 'q'): $q = P(\text{Draw}) + P(\text{P loses})$ $q = 0.15 + 0.05 = 0.20$ So, $q = 0.20$.
We can check that $p + q = 0.80 + 0.20 = 1.00$, which accounts for all possible outcomes for P (win or not win).
Applying the Binomial Probability Formula
This scenario fits the binomial distribution criteria:
- There are a fixed number of trials ($n=3$ matches).
- Each trial is independent (the outcome of one match doesn't affect others).
- Each trial has two possible outcomes: P wins (success) or P does not win (failure).
- The probability of success ($p=0.80$) is the same for each trial.
The binomial probability formula is:
$ P(X=k) = \binom{n}{k} p^k q^{n-k} $Where:
- $n$ = total number of trials
- $k$ = number of successful outcomes
- $p$ = probability of success on a single trial
- $q$ = probability of failure on a single trial
- $\binom{n}{k}$ = the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$
Calculating the Probability
We want to find the probability of P winning exactly 2 matches ($k=2$) out of 3 played ($n=3$).
- Calculate the binomial coefficient $\binom{n}{k}$: $\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1)(1)} = 3$ This means there are 3 possible ways P can win exactly 2 out of 3 matches (e.g., WWL, WLW, LWW).
- Calculate $p^k$: $p^2 = (0.80)^2 = 0.64$
- Calculate $q^{n-k}$: $q^{3-2} = q^1 = (0.20)^1 = 0.20$
- Combine the parts using the formula: $P(X=2) = \binom{3}{2} p^2 q^1$ $P(X=2) = 3 \times (0.80)^2 \times (0.20)^1$ $P(X=2) = 3 \times 0.64 \times 0.20$ $P(X=2) = 3 \times 0.128$ $P(X=2) = 0.384$
- Convert the result to a fraction: $0.384 = \frac{384}{1000}$ Simplify the fraction: $\frac{384}{1000} \div \frac{8}{8} = \frac{48}{125}$
Therefore, the probability of P winning exactly 2 of the 3 matches is $\frac{48}{125}$.
Important Questions from Probability
Two dice are thrown simultaneously and the sum of the numbers appearing on them is noted. What is the probability that the sum is 12?
If a box contains 3 white cushions, 4 red cushions and 5 blue cushions, what is the probability of selecting a white or blue cushion?
A. 2/3
B. 3/4
C. 1/4
D. 1/9Statements followed by some conclusions are given below.
Statements:
1. A bag has 2 white, 3 black, 4 red and 6 green balls.
2. 1 ball selected at random from the bag.
Conclusions:
I. The probability that a black ball is selected is 1/5
II. The probability that a red ball is selected is 6/15
Find which of the conclusions logically follows from the given statement
A. Only conclusion I follows.
B. Only conclusion II follows.
C. Both I and II follow.
D. Neither I nor II follows.
In a shooting test, the probabilities of hitting the target are 1/2 for A, 2/3 for B and 3/4 for C. If they fire at the same target, what is the probability that only one of them hits the target?
A bag contains balls numbered from 1 to 42. One ball is drawn at random from these balls. The probability that its number is a multiple of 7 or 8 is:
