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Probability Workshop — Part I

At the mere mention of probability on the GRE students’ eyes glaze over and all conversation stops, save perhaps for the hushed admission that they don’t like probability.

While probability can be very difficult—and many of the publishers give it short shrift (so there are a lack of practice problems)—keeping a few concepts in mind will make this tricky concept seem somewhat less daunting.

First off we need to know the difference between dependent and independent probability. Today, let’s focus on independent probability.

INDEPENDENT PROBABILITY

If two events have nothing to do with each other then they are independent (makes sense.) For instance, if I toss a fair-sided coin (on average it will come out heads 50% of the time) twice, then the result of one toss does not affect the other toss. So if I get heads on the first toss, then there is still a 50% chance that I will get heads on the next toss.

To illustrate how this concept isn’t always treated rationally, imagine that you have tossed 8 heads in a row. What is the probability that you will toss heads on the 9th toss? It seems very unlikely that you will toss yet another heads. But again, the 9th toss has nothing to do with what happened on the first eight tosses—the chance of tossing a heads is still 50%. Still, most of us would gamble against heads happening again.

To find the probability of a series independent events happening, we find the probability of each event happening and then multiply those probabilities together.

So let’s go back to our streak of 8 heads in a row. To toss this we will have to multiply the probability of getting heads on one toss (50%) eight times. Let’s convert 50% to ½, for ease of calculation. We get (1/2)^8  =1/256 (not a very likely scenario.)

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5 Responses to Probability Workshop — Part I

  1. Prashant P May 21, 2017 at 4:54 am #

    Hey Chris asking you about another magoosh blog post on probability here – as I dont see the feature to ask question / leave a reply there:

    Solution to question 5 on this page inexplicably does not discount overlap [advanced degree + female], while solution to question 2 discounts overlap through not so well explained logic:
    https://magoosh.com/gmat/2012/gmat-math-probability-rules/

    Could you kindly elaborate.

    Thanks so much!

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert May 21, 2017 at 4:09 pm #

      Hi Prashant,

      So, the approaches to these two problems are different. In Question 5, there is no double-counting, so no need to discount the overlap. The two categories are female and advanced degrees. So, we first capture females (some who may have advanced degrees and some who may not), which is 100. Then, we capture males with advanced degrees, which is 60. Boom 160! So, 160 fulfill these criteria out of the 400. And, we never double counted the females with advanced degrees, so no need to discount the overlap. I hope this help! 😀

  2. Prashant P May 21, 2017 at 4:41 am #

    Nice Post Chris!

    I think in these lines – what most people see is the probablity of getting 9 heads in a row is 1/(2^9) = 1/512 – rather than probability of head showing up on 9th toss 1/2. It would seem logical to to bet against 9th head – but I am just a confused sould – can you pl advis why is it wrong to think like most people?

    “What is the probability that you will toss heads on the 9th toss? It seems very unlikely that you will toss yet another heads. But again, the 9th toss has nothing to do with what happened on the first eight tosses—the chance of tossing a heads is still 50%. Still, most of us would gamble against heads happening again.”

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert May 21, 2017 at 3:48 pm #

      Hi Prashant,

      Thank you for your kind words! So, the key here is that each coin toss is independent from the toss before it. If the question asks us solely about the probability of flipping heads on the 9th toss, we have to look at the 9th toss. At the end of the day, the fair coin has 50% of heads, and 50% chance of tails. Now, if the question asks us for the probability of a series of tosses, then we need to take into account the probability that this series of tosses will all happen! This is of course much less likely because you’re betting on a number of rounds, rather than a single round. I hope this help! 🙂

  3. Kai July 4, 2016 at 12:41 pm #

    Hi Chris,

    I always thought what was missing from many probability introductions is the fact that independent means inclusive and exclusive means dependent. Whenever I think about this fact I initially get confused again, until it slowly comes back to memory.

    Independent means that other events can either happen or not happen. On the other hand, mutually exclusive means that events are dependent, so force other events to happen or preclude other events from happening.

    Maybe you could add this right at the top of the explanation, where I feel it belongs.


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