Question:
A discrete random variable \(X\) has the distribution \(B(15,p)\). Given that
\[
\operatorname{Var}(X)=3.15,
\]
then the two possible values of \(p\) are
A discrete random variable \(X\) has the distribution \(B(15,p)\). Given that
\[
\operatorname{Var}(X)=3.15,
\]
then the two possible values of \(p\) are
Show Hint
For binomial distribution \(B(n,p)\), always remember:
\[
\text{Mean}=np
\]
and
\[
\text{Variance}=np(1-p).
\]
Updated On: Jun 26, 2026
- \(0.1\)
- \(0.1,0.9\)
- \(0.4,0.6\)
- \(0.3,0.7\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Recall variance of binomial distribution.
For a binomial distribution \[ B(n,p), \] the variance is \[ npq \] where \[ q=1-p \]
Step 2: Substitute the given values.
Here, \[ n=15 \] and \[ \operatorname{Var}(X)=3.15 \] So, \[ 15p(1-p)=3.15 \]
Step 3: Simplify the equation.
Dividing both sides by \(15\), \[ p(1-p)=\frac{3.15}{15} \] \[ p(1-p)=0.21 \]
Step 4: Form a quadratic equation.
\[ p-p^2=0.21 \] \[ p^2-p+0.21=0 \]
Step 5: Solve the quadratic equation.
\[ p^2-p+0.21=0 \] This can be written as \[ p^2-p+\frac{21}{100}=0 \] Using factorization, \[ (p-0.3)(p-0.7)=0 \]
Step 6: Find the possible values of \(p\).
Therefore, \[ p=0.3 \] or \[ p=0.7 \] Both values are valid because probabilities lie between \(0\) and \(1\).
Step 7: Final conclusion.
Hence, the two possible values of \(p\) are \[ \boxed{0.3,0.7} \]
For a binomial distribution \[ B(n,p), \] the variance is \[ npq \] where \[ q=1-p \]
Step 2: Substitute the given values.
Here, \[ n=15 \] and \[ \operatorname{Var}(X)=3.15 \] So, \[ 15p(1-p)=3.15 \]
Step 3: Simplify the equation.
Dividing both sides by \(15\), \[ p(1-p)=\frac{3.15}{15} \] \[ p(1-p)=0.21 \]
Step 4: Form a quadratic equation.
\[ p-p^2=0.21 \] \[ p^2-p+0.21=0 \]
Step 5: Solve the quadratic equation.
\[ p^2-p+0.21=0 \] This can be written as \[ p^2-p+\frac{21}{100}=0 \] Using factorization, \[ (p-0.3)(p-0.7)=0 \]
Step 6: Find the possible values of \(p\).
Therefore, \[ p=0.3 \] or \[ p=0.7 \] Both values are valid because probabilities lie between \(0\) and \(1\).
Step 7: Final conclusion.
Hence, the two possible values of \(p\) are \[ \boxed{0.3,0.7} \]
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