One book is selected at random. What is probability of getting a social studies book?

\frac{1}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{2}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{1}{3}$
Total science books: $3$
Total Social science books: $3$
Total math books: $1$
Total Langyage books: $2$
Total number of books in bag: $3 + 3 + 1 + 2 = 9$
Required probability:
$P = \frac{3}{9} = \frac{1}{3}$

Suggest Corrections

Similar questions

Q. One book is selected at random. What is the probability of getting a science book?

There are 4 torn books in a lot consisting of 10 books. If 5 books are seleceted at random what is the probability of presence of 2 torn books among the selected?

Q. There are two coins, one unbiased with probability

\frac{1}{2}

of getting heads and the other one is biased with probability

\frac{3}{4}

of getting heads. A coin is selected at random and tossed. It shows heads up. Then the probability that the unbiasedcoin was selected is

Q. In a book of 500 pages , there are 500 misprinting errors, find the probability of at most three misprinting 3 pages selected at random from the book.

A book has pages numbered from 1 to 100. If a page is chosen at random, what is the probability that the sum of the digits on the page is 9?

Join BYJU'S Learning Program

One book is selected at random. What is probability of getting a social studies book?

The correct option is A 13Total science books: 3Total Social science books: 3Total math books: 1Total Langyage books: 2Total number of books in bag: 3+3+1+2=9Required probability:P=39=13

The correct option is A $\frac{1}{3}$
Total science books: $3$
Total Social science books: $3$
Total math books: $1$
Total Langyage books: $2$
Total number of books in bag: $3 + 3 + 1 + 2 = 9$
Required probability:
$P = \frac{3}{9} = \frac{1}{3}$