Video Transcript
Mason spins two four-sided spinners
which are both numbered from one to four. Let 𝐴 be the event that he spins
two numbers with a product of four, and let 𝐵 be the event that he spins two even
numbers. There are four parts to this
question. Find the probability of 𝐴, giving
your answer as a fraction in its simplest form. Find the probability of 𝐵, giving
your answer as a fraction in its simplest form. Find the probability of 𝐴
intersection 𝐵, giving your answer as a fraction in its simplest form. And finally, determine the value of
the probability of 𝐴 union 𝐵, giving your answer as a fraction in its simplest
form.
We will begin by clearing some
space whilst recalling we need to simplify all of our answers. We are told in the question that
Mason is spinning two four-sided spinners that are numbered from one to four. This means that there is 16
possible combinations. These include a one on both
spinners all the way up to a four on both spinners.
We are told that 𝐴 is the event
that he spends two numbers with a product of four. The word product means multiply or
times. We know that one multiplied by four
is four. Two multiplied by two is equal to
four. And four multiplied by one is also
equal to four. There are three possible ways that
Mason can spin the two spinners such that the numbers have a product of four. The probability of 𝐴 is therefore
equal to three out of 16 or three sixteenths. Since the numerator and denominator
have no common factor apart from one, this fraction is in its simplest form.
We are also told that 𝐵 is the
event that Mason spins two even numbers. This occurs when he spends a two
and a two, a two and a four, a four and a two, or a four and a four. There are four possible ways of
spinning two even numbers. Therefore, the probability of 𝐵 is
equal to four sixteenths. Both the numerator and denominator
here are divisible by four. The fraction four sixteenths in its
simplest form is therefore one-quarter. This is the probability that Mason
spends two even numbers.
The third part of our question asks
us to calculate the probability of 𝐴 intersection 𝐵. This is the probability that both
𝐴 and 𝐵 occur. Mason needs to spin two numbers
with a product of four, and they must both be even. The only way this is possible is
when he rolls two twos, as two multiplied by two is four and the number two is
even. The probability of 𝐴 intersection
𝐵 is therefore equal to one sixteenth.
The final part of our question asks
us to determine the value of the probability of 𝐴 union 𝐵. This is the probability that 𝐴
occurs or 𝐵 occurs or they both occur. One way of calculating this is
using the addition rule of probability, which states that the probability of 𝐴
union 𝐵 is equal to the probability of 𝐴 plus the probability of 𝐵 minus the
probability of 𝐴 intersection 𝐵.
In this question, we have already
calculated the probability of 𝐴, the probability of 𝐵, and the probability of 𝐴
intersection 𝐵. The probability of 𝐴 union 𝐵 is
therefore equal to three sixteenths plus four sixteenths minus one sixteenth. This is equal to six
sixteenths. We can check this by looking at the
list of possible outcomes. Six of the 16 outcomes were in
either event 𝐴 or event 𝐵 or both. As both six and 16 are divisible by
two, we can simplify this fraction to three-eighths. The value of the probability of 𝐴
union 𝐵 is three-eighths.
And the four answers to the
question are three sixteenths, one-quarter, one sixteenth, and three-eighths, where
all four fractions are written in their simplest form.
