Question

The random variable \(X\) has Binomial distribution \[ B(20,0.4) \] Then \[ 5-5P(X\geq 2)= \] is equal to:

• A. \(62\left(\dfrac{2}{5}\right)^{19}\)
• B. \(43\left(\dfrac{3}{5}\right)^{19}\)
• C. \(1+23\left(\dfrac{3^{19}}{5^{20}}\right)\)
• D. \(1+62\left(\dfrac{2^{19}}{5^{20}}\right)\)

Hint:

For binomial distribution, \[ P(X\geq r) = 1-\sum_{k=0}^{r-1}P(X=k) \] Always convert cumulative probabilities into simpler complementary probabilities whenever possible.

Answer 1

The Correct Option is B

Step 1: Write the binomial parameters.
$ X \sim B(20, 0.4) $, so $ n = 20 $, $ p = \dfrac{2}{5} $, $ q = 1 - p = \dfrac{3}{5} $. We need $ 5 - 5P(X \geq 2) $.
Step 2: Rewrite the expression using the complement.
\[ 5 - 5P(X \geq 2) = 5[1 - P(X \geq 2)] = 5[P(X = 0) + P(X = 1)] \] This uses the fact that $ P(X \geq 2) = 1 - P(X=0) - P(X=1) $.
Step 3: Compute $ P(X = 0) $.
\[ P(X=0) = \binom{20}{0}\left(\frac{2}{5}\right)^0\left(\frac{3}{5}\right)^{20} = \left(\frac{3}{5}\right)^{20} \]
Step 4: Compute $ P(X = 1) $.
\[ P(X=1) = \binom{20}{1}\left(\frac{2}{5}\right)^1\left(\frac{3}{5}\right)^{19} = 20 \cdot \frac{2}{5} \cdot \left(\frac{3}{5}\right)^{19} = 8\left(\frac{3}{5}\right)^{19} \]
Step 5: Add and factor.
\[ P(X=0) + P(X=1) = \left(\frac{3}{5}\right)^{20} + 8\left(\frac{3}{5}\right)^{19} = \left(\frac{3}{5}\right)^{19}\left(\frac{3}{5} + 8\right) = \left(\frac{3}{5}\right)^{19} \cdot \frac{43}{5} \]
Step 6: Multiply by 5 and state the answer.
\[ 5[P(X=0)+P(X=1)] = 5 \cdot \frac{43}{5} \cdot \left(\frac{3}{5}\right)^{19} = 43\left(\frac{3}{5}\right)^{19} \] \[ \boxed{43\left(\frac{3}{5}\right)^{19}} \]