Independent solution

How to solve this Binomial Distribution question

Setup

Setup

Let p be the claim probability and let q=1-p. Translate each stopping event into a Bernoulli-sequence probability.

q=1−pq=1-p
Pr⁡(first claim at trial 6)=q5p\Pr(\text{first claim at trial }6)=q^5p
Pr⁡(second claim at trial 8)=(71)p2q6\Pr(\text{second claim at trial }8)=\binom{7}{1}p^2q^6

Model

Model

Equate the probabilities and cancel factors that are positive for an interior probability.

q5p=7p2q6q^5p=7p^2q^6
1=7pq=7p(1−p)1=7pq=7p(1-p)

Compute

Compute

Solve the resulting quadratic and apply the stated restriction that p is below one half.

7p2−7p+1=07p^2-7p+1=0
p=7±2114p=\frac{7\pm\sqrt{21}}{14}
p=7−2114=0.1726731646…p=\frac{7-\sqrt{21}}{14}=0.1726731646\ldots

Answer

Answer

The smaller root rounds to 0.173 and matches choice C.

p≈0.173(C)\boxed{p\approx0.173\quad\text{(C)}}