Question

An exam can be given maximum four times. The probability of selection in first attempt is 1/24 and it increases by 50% for every next attempt. The probability of selection is ______.Given that individual can give only one successful attempt.

• A. 0.2
• B. 0.3
• C. 0.4
• D. 0.5

Answer 1

The correct option is B 0.3

Let $P_1,P_2,P_3,P_4$ be probability of selection in $1^{st},2^{nd},3^{rd}\&4^{th}$ attempt respectively.
Now,

$P_1=\frac{1}{24};P_2=\frac{1}{24}[1+0.5]$

$P_2=\frac{1}{24}\times \frac{3}{2}$

$P_3=\frac{1}{24}\times \frac{3}{2}[1+0.5]=\frac{1}{24}\times {\left(\frac{3}{2}\right)}^2$

$P_4=\frac{1}{24}\times {\left(\frac{3}{2}\right)}^3$


Now let $A_i$ be selection in $i^{th}$ attempt & $\overline{A_i}$ be unsuccessful attempt,

So,

$P_{selection}=A_1+\overline{A_1}{A}_2+\overline{A_1}\overline{A_2}{A}_3+\overline{A_1}\overline{A_2}\overline{A_3}{A}_4$

$=\frac{1}{24}+\frac{23}{24}\times \frac{1}{24}\times \frac{3}{2}+\frac{23}{24}(1-\frac{3}{48})\times \frac{1}{24}\times {\left(\frac{3}{2}\right)}^2+\frac{23}{24}(1-\frac{3}{48})(1-\frac{9}{96})\frac{1}{24}\times {\left(\frac{3}{2}\right)}^3=0.3$