- A$\frac{105}{16}$
- ✓$\frac{77}{36}$
- C$\frac{7}{16}$
- D$\frac{49}{16}$
$n =7 \quad p = p$
given
$P(x=3)=5 P(x=4)$
${ }^{7} C_{3} \times p^{3}(1-p)^{4}=5^{7} C_{4} p^{4}(1-p)^{3}$
$\frac{{ }^{7} C_{3}}{5 \times{ }^{7} C_{4}}=\frac{p}{1-p}$
$1- p =5 p$
$6 p =1$
$p=\frac{1}{6} \Rightarrow q=\frac{5}{6}$
$n =7$
Mean $= np =7 \times \frac{1}{6}=\frac{7}{6}$
Var $= npq =7 \times \frac{1}{6} \times \frac{5}{6}=\frac{35}{36}$
Sum
$=\frac{7}{6}+\frac{35}{36}$
$=\frac{42+35}{36}$
$=\frac{77}{36}$
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A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two consecutive digits are identical, is:
$\frac{1}{5}$
$\frac{1}{5}\big(\frac{9}{10}\big)^3$
$\big(\frac{3}{5}\big)^4$
$\text{None of these}$