Math  /  Data & Statistics

Questionminules.
1. Peter. Tom, and Carl simultancously shoot at a target. Peter hits the target with a probability of 1/21 / 2. Tho with a probability of 2/32 / 3, and Carl with a probability of 3/43 / 4. a) Assume that only one of them hit the target. What is the probability that it was Carl? b) What is the most likely number of hits on the larget? Determine the expected vilue of the uumber hits.

Studdy Solution
Calculate the expected value of the number of hits:
E(hits)=0×124+1×14+2×1124+3×14 E(\text{hits}) = 0 \times \frac{1}{24} + 1 \times \frac{1}{4} + 2 \times \frac{11}{24} + 3 \times \frac{1}{4}
E(hits)=0+14+2224+34 E(\text{hits}) = 0 + \frac{1}{4} + \frac{22}{24} + \frac{3}{4}
E(hits)=624+2224+1824 E(\text{hits}) = \frac{6}{24} + \frac{22}{24} + \frac{18}{24}
E(hits)=4624=2312 E(\text{hits}) = \frac{46}{24} = \frac{23}{12}
The probability that it was Carl who hit the target given only one hit is 12 \frac{1}{2} .
The most likely number of hits is 2.
The expected value of the number of hits is 2312 \frac{23}{12} .

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord