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 given distribution.
Given \[ X\sim B(20,0.4) \] Hence, \[ n=20,\quad p=0.4=\frac{2}{5} \] Therefore, \[ q=1-p=1-\frac{2}{5}=\frac{3}{5} \]

Step 2: Express \(P(X\geq 2)\).
We know that \[ P(X\geq 2)=1-P(X=0)-P(X=1) \] Therefore, \[ 5-5P(X\geq 2) = 5\left[1-P(X\geq 2)\right] \] So, \[ 5-5P(X\geq 2) = 5[P(X=0)+P(X=1)] \]

Step 3: Calculate \(P(X=0)\).
Using binomial formula, \[ P(X=0) = {}^{20}C_0 \left(\frac{2}{5}\right)^0 \left(\frac{3}{5}\right)^{20} \] Thus, \[ P(X=0) = \left(\frac{3}{5}\right)^{20} \]

Step 4: Calculate \(P(X=1)\).
Again, \[ P(X=1) = {}^{20}C_1 \left(\frac{2}{5}\right)^1 \left(\frac{3}{5}\right)^{19} \] \[ = 20\times \frac{2}{5} \left(\frac{3}{5}\right)^{19} \] \[ = 8\left(\frac{3}{5}\right)^{19} \]

Step 5: Add the probabilities.
Now, \[ P(X=0)+P(X=1) = \left(\frac{3}{5}\right)^{20} + 8\left(\frac{3}{5}\right)^{19} \] Taking \[ \left(\frac{3}{5}\right)^{19} \] common, \[ = \left(\frac{3}{5}\right)^{19} \left(\frac{3}{5}+8\right) \] \[ = \left(\frac{3}{5}\right)^{19} \left(\frac{3+40}{5}\right) \] \[ = \frac{43}{5} \left(\frac{3}{5}\right)^{19} \] Therefore, \[ 5[P(X=0)+P(X=1)] = 43\left(\frac{3}{5}\right)^{19} \]

Step 6: Final conclusion.
Hence, \[ \boxed{ 43\left(\frac{3}{5}\right)^{19} } \]