Question

A coin whose faces are marked 3 and 5 is tossed 4 times; what are the odds against the sum of the numbers thrown being less than 15?

• A. The odds against are $5:11$
• B. The odds against are $11:5$
• C. The odds against are $16:5$
• D. The odds against are $5:16$

Answer 1

Answer: (B)

The odds against are $11:5$

Total number of cases $=2^4=16$
Favourable cases for sum less than $15$ are
(i) All comes with $3$ up $3,3,3,3$
(II)Three coins with $3$ up and one with $5$ up i.e. $(5,3,3,3),(3,5,3,3),(3,3,5,3),(3,3,3,5)$
So, number of favourable cases $=5$
So, probability $=\frac{5}{16}$
Odds in favour are $5:16$
Odds against are $11:16$