A bag contains 21 toys numbered 1 to 21. A toy is drawn and then another toy is drawn without replacement.
Find the probability that both toys will show even numbers.
A. $$\frac{{5}}{{21}}$$
B. $$\frac{{9}}{{42}}$$
C. $$\frac{{11}}{{42}}$$
D. $$\frac{{4}}{{21}}$$
Answer: Option B
Solution(By Examveda Team)
The probability that first toy shows the even number $$ = \frac{{10}}{{21}}$$Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.
Hence, probability that second toy shows the even number $$ = \frac{{9}}{{20}}$$
Required probability,
$$\eqalign{ & = \left( {\frac{{10}}{{21}}} \right) \times \left( {\frac{9}{{20}}} \right) \cr & = \frac{9}{{42}} \cr} $$
Related Questions on Probability
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{8}}{{15}}$$
D. $$\frac{{9}}{{20}}$$
A. $$\frac{{10}}{{21}}$$
B. $$\frac{{11}}{{21}}$$
C. $$\frac{{2}}{{7}}$$
D. $$\frac{{5}}{{7}}$$
A. $$\frac{{1}}{{3}}$$
B. $$\frac{{3}}{{4}}$$
C. $$\frac{{7}}{{19}}$$
D. $$\frac{{8}}{{21}}$$
E. $$\frac{{9}}{{21}}$$
What is the probability of getting a sum 9 from two throws of a dice?
A. $$\frac{{1}}{{6}}$$
B. $$\frac{{1}}{{8}}$$
C. $$\frac{{1}}{{9}}$$
D. $$\frac{{1}}{{12}}$$
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