Question 23

A bag contains 5 white and 3 black balls, and 4 are successively drawn out and not replaced. What’s the chance of getting different colours alternatively?

Solution

Probability of alternate balls when first ball is black = $$\frac{5}{8}\times\ \frac{3}{7}\times\ \frac{4}{6}\times\ \frac{2}{5}=\frac{1}{14}$$

Probability of alternate balls when first ball is white = $$\frac{3}{8}\times\ \frac{5}{7}\times\ \frac{2}{6}\times\ \frac{4}{5}=\frac{1}{14}$$

Overall Probability = $$\frac{1}{14}+\ \frac{1}{14}=\frac{1}{7}$$

Create a FREE account and get:

All Quant Formulas and shortcuts PDF
40+ previous papers with solutions PDF
Top 500 MBA exam Solved Questions for Free

CAT Quant Questions | CAT Quantitative Ability

CAT Algebra Questions CAT Data Sufficiency Questions CAT Time, Distance and Work Questions CAT Number Systems Questions CAT Linear Equations Questions

CAT DILR Questions | LRDI Questions For CAT

CAT DI with connected data sets Questions CAT Routes And Networks Questions CAT Special Charts Questions CAT Table with Missing values Questions CAT Selection With Condition Questions

CAT Verbal Ability Questions | VARC Questions For CAT

CAT Vocabulary Questions CAT Miscellaneous Questions CAT Odd One Out Questions CAT Verbal Ability Questions CAT Critical Reasoning Questions

Follow us on

Boost your Prep!

Download App