Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the probability distribution of the number of successes in two tosses of a die where a success is defined as (i) number greater than 4

Q. 5  Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as

             (i) number greater than 4

Answers (1)

best_answer

When a die is tossed twice , total outcomes = 36

Number less than or equal to 4 in both toss  :  P(X=0)=\frac{4}{6} \times \frac{4}{6}=\frac{4}{9}

Number less than or equal to 4 in first toss and number more than or equal to 4 in second toss + Number less than or equal to 4 in second toss and number more than or equal to 4 in first toss: P(X=1)=\frac{4}{6} \times \frac{2}{6}+ \frac{2}{6}\times \frac{4}{6}=\frac{4}{9}

Number less than 4 in both tosses : P(X=2)=\frac{2}{6} \times \frac{2}{6}= \frac{1}{9}

Probability distribution is as :

X 0 1 2
P(X) \frac{4}{9} \frac{4}{9} \frac{1}{9}

 

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 Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years
anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question