Video: CBSE Class X • Pack 5 • 2014 • Question 13

CBSE Class X • Pack 5 • 2014 • Question 13

03:06

Video Transcript

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 four.

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 three-quarters.

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