Question

26. Convert each of the following probabilities to odds: (a) The probability of getting two heads in six flips of a balan (b) The probability of drawing a one-eyed jack from (c) The probability that the last digit of a car's license plate is a 3, 4 ced coin is 8 deck of 52 playing cards is 5, 6,7,8, or 9 is o an ordinary

Answer 1

a) Let A be the event of getting 2 heads in six flips of a balanced coin.

the probability of getting 2 heads in six flips of a balanced coin is

$P(A)=\frac{15}{64}$

the probability of not getting 2 heads in six flips of a balanced coin is

$P(A^c)=1-P(A)=1-\frac{15}{64}=\frac{49}{64}$

the odds of getting 2 heads in six flips of a balanced coin is

$\text{Odds(A)}=\frac{P(A)}{P(A^c)}=\frac{P(A)}{1-P(A)}=\frac{\frac{15}{64}}{\frac{49}{64}}=15:49$

ans: odds of getting 2 heads in six flips of a balanced coin is 15:49

b) Let B be the event of drawing a one-eyed jack from an ordinary deck of 52 playing cards.

The probability of drawing a one-eyed jack from an ordinary deck of 52 playing cards is

$P(B)=\frac{2}{52}$

the odds of drawing a one-eyed jack from an ordinary deck of 52 playing cards is

$\text{Odds(B)}=\frac{P(B)}{1-P(B)}=\frac{\frac{2}{52}}{1-\frac{2}{52}}=2:50=1:25$

ans: the odds of drawing a one-eyed jack from an ordinary deck of 52 playing cards is 1:25

c) Let C be the event that the last digit of a car's license plate is a 3,4,5,6,7,8, or 9.

the probability of  the last digit of a car's license plate is a 3,4,5,6,7,8, or 9 is

$P(C)=\frac{7}{10}$

the odds of the last digit of a car's license plate is a 3,4,5,6,7,8, or 9 is

$\text{Odds(C)}=\frac{P(C)}{1-P(C)}=\frac{\frac{7}{10}}{1-\frac{7}{10}}=7:3$

ans: the odds of the last digit of a car's license plate is a 3,4,5,6,7,8, or 9 is 7:3