A die is tossed three times, What is the probability of getting an odd number at least once ?

Question

Question

This question was previously asked in

DFCCIL Executive S&T 2018 Official Paper

Answer 1

A die is having six sides with six different numbering

Even number in the die are {2,4,6}

The odd number in the die are {1,3,5}

\(Probability~of~an~Event =\frac{Number~of~desired~Events }{Total~number~of~Events}\)

The probability of occurring an odd number is = 3/6 = 1/2

The probability of occurring an even number is = 3/6 = 1/2

When die is tossed three times and getting an odd number at least once is given by

p(at least one odd number) = 1 - p(no odd number in all three tosses)

\(p(at~least~one~odd~number) =1- \frac{1}{8}\)

p(no odd number in all three tosses) = p(even) × p(even) × p(even)

\(p(no~odd~number~in~all~three~tosses) = \frac{1}{2}×\frac{1}{2}×\frac{1}{2}\)

\(p(no~odd~number~in~all~three~tosses) = \frac{1}{8}\)

\(p(at~least~one~odd~number) =\frac{7}{8}\)

Hence option (2) is correct