Rahim flips two different coins
simultaneously. Find the probability that at least
one lands on tails.
For this type of question, one
technique we can use is to draw a sample space diagram. And this will allow us to observe
all of the different outcomes of Rahim flipping the two coins. For the purpose of this question,
we’ll assume that each coin can only land on heads and tails. We’re also going to assume that
both of our coins are fair. Which means that it’s equally
likely they will land on heads and tails.
Here, we see the possible outcomes
for our first coin, which we’ll call coin one. For our sample space diagram, we
can also draw in the possible outcomes for coin two, which is also heads and
tails. Arranging these in a table allows
us to fill in all the possible outcomes.
For example, if Rahim flips heads
on coin one and heads on coin two, then the outcome is heads and heads. Likewise, if Rahim flips heads on
coin one and tails on coin two, then the outcome is heads and tails. Following this same pattern, we see
the remaining two outcomes are tails and heads. And finally, tails and tails.
We have now found four total
outcomes for Rahim flipping two different coins. Now, the question asked us to find
the probability that at least one of the coins lands on tails. Looking at our outcomes, we see
that this occurs three different times. First, in the case where coin one
lands on tails. Next, in the case where coin two
lands on tails. And finally, in the case where both
coin one and coin two land on tails, since this still satisfies the condition that
at least one of them is landing on tails. We now see that we have three of
four possible outcomes which satisfy the condition.
So earlier on, we noted that both
of the coins are fair. This is important because we said
that they are equally likely to land on heads and tails. It therefore follows that each of
our four different outcomes are also equally likely to occur. Because of this, we can say the
following: if event 𝑋 is defined as the event where Rahim flips at least one tails,
then the probability of 𝑋 is equal to the number of outcomes which satisfy 𝑋
divided by the total number of outcomes in the sample space.
Looking at our sample space
diagram, we know that 𝑛 𝑆, or the total number of outcomes in the sample space, is
four. We have also shown that 𝑛 𝑋, the
number of outcomes which satisfy the event of flipping at least one tails, is one,
two, three. This allows us to say that the
probability that at least one of Rahim’s coins lands on tails is three over
Since this fraction is already in
simplified form, we have therefore answered the question. And we found that the probability
that at least one of Rahim’s coins lands on tails is three over four or