Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Suppose that 90 % of people are right -handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Q. 4  Suppose that 90^{o}/_{o} of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Answers (1)

best_answer

90^{o}/_{o} of people are right-handed.

P(right-handed)=\frac{9}{10}

P(left-handed)=q=1-\frac{9}{10}=\frac{1}{10}

at most 6 of a random sample of 10 people are right-handed.

the probability that more than  6 of a random sample of 10 people are right-handed is given by,

                                                        \sum_{T}^{10} ^{10}C_r P^{r} q^{10-r}

                                                       =\sum_{T}^{10} ^{10}C_r\frac{9}{10}^r .(\frac{1}{10})^{10-r}

the probability that at most 6 of a random sample of 10 people are right-handed is given by

                                                      =1-\sum_{T}^{10} ^{10}C_r .\frac{9}{10}^r .(\frac{1}{10})^{10-r}

Posted by

seema garhwal

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question