Probability: questions with answers

The bag has $1 + 9 = 10$ balls in all, so there are $10$ equally likely outcomes. The favourable outcomes are the red ones, and the red balls number $1$. So $P(\text{red}) = \dfrac{ 1 }{ 10 }$. Dividing by the $9$ black balls instead of the total gives $\dfrac{ 1 }{ 9 }$. That compares red with black only, and forgets that red balls belong to the total too. $\dfrac{ 9 }{ 10 }$ is the probability of drawing a black ball, not a red one. And $9$ is only a count, not a probability.

Every ball is either red or blue, so these events are complementary. Thus $P(\text{blue}) = 1 - \dfrac{ 2 }{10} = \dfrac{ 8 }{10}$. Since each ball is equally likely to be drawn, $\dfrac{\text{number of blue balls}}{ 30 } = \dfrac{ 8 }{10}$. Therefore, the number of blue balls is $30 \times \dfrac{ 8 }{10} = 24$. Adding the red probability to one incorrectly gives $36$ blue balls, more than the total. $6$ is the number of red balls, and $30$ is the total number of balls.

The possible outcomes are the $27$ balls he played, and $12$ of them were hit for a boundary, so the probability is $\dfrac{ 12 }{ 27 } = \dfrac{ 4 }{ 9 }$. The denominator of a probability is always the total number of outcomes, not the number of unfavourable ones. $\dfrac{ 4 }{ 5 }$ is the ratio of boundaries to other balls, so it is not a probability. Reversed, $\dfrac{ 5 }{ 4 }$ is even greater than $1$. $\dfrac{ 5 }{ 9 }$ is the probability that he did NOT hit a boundary.

One card is missing, so the pack holds $52 - 1 = 51$ cards and those $51$ are the only possible outcomes. $13$ of them answer the description, so the probability is $\dfrac{ 13 }{ 51 }$. Leaving the denominator at $52$ gives $\dfrac{ 1 }{ 4 }$, counting a card that is no longer there. $\dfrac{ 26 }{ 51 }$ comes from $26$ - a true count of the deck, but of a suit where a colour was meant, or of a colour where a suit was. $\dfrac{ 38 }{ 51 }$ is the probability of the opposite event.

Removing a whole suit takes $13$ cards away, so the pack in front of you holds $52 - 13 = 39$ cards - and those $39$ are the only possible outcomes. $26$ of them are a black card, so the probability is $\dfrac{ 26 }{39} = \dfrac{ 2 }{ 3 }$. Leaving the denominator at $52$ gives $\dfrac{ 1 }{ 2 }$, which counts cards that are no longer in the pack. $\dfrac{ 1 }{ 3 }$ is the probability that the card is not a black card, and $\dfrac{ 9 }{ 13 }$ forgets that one card of that kind went with the suit.

The face cards are the kings, queens and jacks - $12$ in the deck, $6$ in each colour, $3$ in each suit - and the aces are not among them. So the red face cards takes $6$ cards away and leaves $46$, of which $4$ are an ace: $P = \dfrac{ 4 }{ 46 } = \dfrac{ 2 }{ 23 }$. Leaving the denominator at $52$ gives $\dfrac{ 1 }{ 13 }$, and counting the aces as face cards changes BOTH counts and gives $\dfrac{ 1 }{ 22 }$. $\dfrac{ 21 }{ 23 }$ is the probability of the opposite event.

The $16$ bad eggs are $\dfrac{ 2 }{ 5 }$ of the whole lot, so the lot holds $\dfrac{ 16 \times 5 }{ 2 } = 40$ eggs, and the good ones number $40 - 16 = 24$. Writing $\dfrac{ 16 }{x} = \dfrac{ 2 }{ 5 }$ with $x$ the number of GOOD eggs leaves the bad eggs out of the total. It returns $40$, the size of the whole lot, not of its good part. And $3$ is only the numerator of $\dfrac{ 3 }{ 5 }$, the probability that an egg is good. A probability is a share of the whole, becoming a count only after multiplying by the total.

The pen is drawn from the whole lot, so there are $108$ equally likely outcomes, and $108 - 48 = 60$ of the pens are good: $P = \dfrac{ 60 }{ 108 } = \dfrac{ 5 }{ 9 }$. Dividing the good pens by the bad ones gives $\dfrac{ 5 }{ 4 }$. This leaves the good pens out of the denominator, and the denominator of a probability is always ALL the outcomes. $\dfrac{ 4 }{ 9 }$ is the probability that she does not buy it, and $60$ is a count of pens, not a probability.

The ratio splits the carton into $2 + 5 = 7$ equal parts, so one part is $\dfrac{ 105 }{ 7 } = 15$ balls and the blue balls are $2$ parts: $2 \times 15 = 30$. Taking $\dfrac{ 2 }{ 5 }$ of the carton instead gives $42$, using the ratio as though it were the share. This compares the blue balls to the yellow balls, not to the whole carton. $75$ is the number of yellow balls, and $105$ is the whole carton.

The die has $6$ equally likely faces, $1$ to $6$. The primes among them are $2$, $3$ and $5$; the composites are $4$ and $6$; and $1$ is neither, because a prime has exactly two distinct factors while $1$ has only one. So a prime number is shown by $3$ of the $6$ faces, and the probability is $\dfrac{ 3 }{6} = \dfrac{1}{2}$. Counting $1$ in with the primes makes it $4$ faces and gives $\dfrac{2}{3}$ instead. And $3$ on its own is the COUNT of favourable faces: a probability has still to be divided by the $6$ possible outcomes.

Counting both end numbers, the box holds $20 - 3 + 1 = 18$ cards. Of these, $6$ have numbers that are multiples of $3$. Each card is equally likely to be drawn, so the probability is $\dfrac{ 6 }{ 18 } = \dfrac{1}{ 3 }$. Using $20 - 3 = 17$ as the total leaves out one card. The number $6$ alone counts favourable outcomes; it is not the probability. The simplified probability has denominator $3$, not $4$.

A run of whole numbers from $4$ to $40$, counting both ends, holds $40 - 4 + 1 = 37$ numbers. Subtracting the two numbers gives $36$, which counts the gaps between the cards rather than the cards - it is short by exactly one. $40$ would be the answer only if the numbering started at $1$, and $41$ counts one card too many, one place past the last card.

The $6 \times 6$ table holds $36$ ordered outcomes. Six outcomes show the same number on both dice. The other $30$ outcomes show different numbers, split evenly between a first number that is greater and one that is smaller. Here the same number on both dice covers $6$ outcomes, so $P = \dfrac{ 6 }{36} = \dfrac{ 1 }{ 6 }$. Treating a pair and its reverse as one outcome shrinks the table to $21$ entries and gives $\dfrac{ 2 }{ 7 }$ instead. But the two dice can be told apart, so $(2, 5)$ and $(5, 2)$ are two separate outcomes. $\dfrac{ 5 }{ 6 }$ is the probability of the opposite event.

The smallest sum two dice can show is $1 + 1 = 2$ and the largest is $6 + 6 = 12$, so no entry in the $6 \times 6$ table gives a sum equal to $15$. The event is favoured by NO outcome, which makes it impossible: $P = \dfrac{0}{36} = 0$. It is not a small fraction with $36$ underneath - $\dfrac{1}{36}$ would mean one of the thirty-six outcomes does give it, and none does. And $1$ is the value of a CERTAIN event, the opposite extreme.

The sums two dice can show run from $1 + 1 = 2$ to $6 + 6 = 12$, so EVERY one of the $6 \times 6 = 36$ outcomes has a sum at least $2$. The event is favoured by all of them, which makes it certain: $P = \dfrac{36}{36} = 1$. $\dfrac{35}{36}$ would wrongly mean one outcome escapes, but none does. $\dfrac{7}{12}$ is $\dfrac{21}{36}$, left over only if a pair and its reverse are wrongly counted as one outcome. $0$ is the value of an IMPOSSIBLE event, not a certain one.