Answer :
To solve the problem of finding the probability that the number of correct answers [tex]\( X \)[/tex] is fewer than 4 in a scenario where there are [tex]\( n = 6 \)[/tex] trials, each with a probability of success [tex]\( p = 0.35 \)[/tex], we can use the binomial distribution formula.
The binomial distribution provides the probability of having exactly [tex]\( k \)[/tex] successes in [tex]\( n \)[/tex] independent Bernoulli trials with the same probability of success [tex]\( p \)[/tex]. The probability mass function (PMF) for a binomial distribution is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
Where:
- [tex]\(\binom{n}{k} = \frac{n!}{k! (n-k)!}\)[/tex] is the binomial coefficient,
- [tex]\( n = 6 \)[/tex] is the number of trials,
- [tex]\( p = 0.35 \)[/tex] is the probability of success on a single trial,
- [tex]\( k \)[/tex] is the number of successes (correct answers).
We need to find the probability that [tex]\( X \)[/tex] is fewer than 4, which means summing the probabilities for [tex]\( X = 0 \)[/tex], [tex]\( X = 1 \)[/tex], [tex]\( X = 2 \)[/tex], and [tex]\( X = 3 \)[/tex]:
[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]
Now we compute each term separately:
1. [tex]\( P(X = 0) \)[/tex]:
[tex]\[ P(X = 0) = \binom{6}{0} (0.35)^0 (0.65)^6 = 1 \times 1 \times (0.65)^6 \approx 0.1160 \][/tex]
2. [tex]\( P(X = 1) \)[/tex]:
[tex]\[ P(X = 1) = \binom{6}{1} (0.35)^1 (0.65)^5 = 6 \times 0.35 \times (0.65)^5 \approx 0.1618 \][/tex]
3. [tex]\( P(X = 2) \)[/tex]:
[tex]\[ P(X = 2) = \binom{6}{2} (0.35)^2 (0.65)^4 = 15 \times (0.35)^2 \times (0.65)^4 \approx 0.2248 \][/tex]
4. [tex]\( P(X = 3) \)[/tex]:
[tex]\[ P(X = 3) = \binom{6}{3} (0.35)^3 (0.65)^3 = 20 \times (0.35)^3 \times (0.65)^3 \approx 0.3800 \][/tex]
Adding these probabilities:
[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]
[tex]\[ P(X < 4) \approx 0.1160 + 0.1618 + 0.2248 + 0.3800 = 0.8826 \][/tex]
Therefore, the probability that the number of correct answers is fewer than 4 is [tex]\( \boxed{0.8826} \)[/tex].
The binomial distribution provides the probability of having exactly [tex]\( k \)[/tex] successes in [tex]\( n \)[/tex] independent Bernoulli trials with the same probability of success [tex]\( p \)[/tex]. The probability mass function (PMF) for a binomial distribution is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
Where:
- [tex]\(\binom{n}{k} = \frac{n!}{k! (n-k)!}\)[/tex] is the binomial coefficient,
- [tex]\( n = 6 \)[/tex] is the number of trials,
- [tex]\( p = 0.35 \)[/tex] is the probability of success on a single trial,
- [tex]\( k \)[/tex] is the number of successes (correct answers).
We need to find the probability that [tex]\( X \)[/tex] is fewer than 4, which means summing the probabilities for [tex]\( X = 0 \)[/tex], [tex]\( X = 1 \)[/tex], [tex]\( X = 2 \)[/tex], and [tex]\( X = 3 \)[/tex]:
[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]
Now we compute each term separately:
1. [tex]\( P(X = 0) \)[/tex]:
[tex]\[ P(X = 0) = \binom{6}{0} (0.35)^0 (0.65)^6 = 1 \times 1 \times (0.65)^6 \approx 0.1160 \][/tex]
2. [tex]\( P(X = 1) \)[/tex]:
[tex]\[ P(X = 1) = \binom{6}{1} (0.35)^1 (0.65)^5 = 6 \times 0.35 \times (0.65)^5 \approx 0.1618 \][/tex]
3. [tex]\( P(X = 2) \)[/tex]:
[tex]\[ P(X = 2) = \binom{6}{2} (0.35)^2 (0.65)^4 = 15 \times (0.35)^2 \times (0.65)^4 \approx 0.2248 \][/tex]
4. [tex]\( P(X = 3) \)[/tex]:
[tex]\[ P(X = 3) = \binom{6}{3} (0.35)^3 (0.65)^3 = 20 \times (0.35)^3 \times (0.65)^3 \approx 0.3800 \][/tex]
Adding these probabilities:
[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]
[tex]\[ P(X < 4) \approx 0.1160 + 0.1618 + 0.2248 + 0.3800 = 0.8826 \][/tex]
Therefore, the probability that the number of correct answers is fewer than 4 is [tex]\( \boxed{0.8826} \)[/tex].
