Probability Question 52
Question: Two persons each make a single throw with a die. The probability they get equal value is $ p_1 $ . Four persons each make a single throw and probability of three being equal is $ p_2 $ , then
Options:
A) $ p_1=p_2 $
B) $ p_1<p_2 $
C) $ p_1>p_2 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ p_1=\frac{6}{36}=\frac{1}{6} $
To find $ p_2, $ the total number of ways $ =6^{4} $ and since two numbers out of 6 can be selected in $ {}^{6}C_2 $ ways i.e. 15 ways and corresponding to each of these ways, there are 8 ways e.g., $ (1,1,1,2)(1,1,2,1)….. $
Thus favourable ways $ =15\times 8=120 $
Hence $ p_2=\frac{120}{6^{4}}=\frac{5}{54} $ . Hence $ p_1>p_2. $
BETA
