In Magoosh, it,s a hard question not medium.

can you please elaborate this calculating each part? Because, there may be some questions in GRE which needs calculations.

For, the probability of getting two heads exactly, when coin is tossed 5 times,

the no. of favorable outcome= arrangement of 2 H and 3 T(HHTTT)= 5!/(2!*3!)=10
the total no of outcomes for tossing coin 5 times =2^5=32
therefore required probablity=10/32

If we follow same steps for getting exactly 3 heads the result will be same.... am I incorrect????

someone please follow up with subasyahoo. i'd like to to see the calculations as well - for quantity A and quantity B. This will be very helpful.

Given that When a coin is flipped, the probability of getting heads is 0.5, and the probability of getting tails is 0.5 A coin is flipped 5 times.

Coin is tossed 5 times => Total number of cases = \(2^5\) = 32

Quantity A: Probability of getting exactly 2 Heads

We need to get 2 heads out of 5 tosses.

We have 5 places to fill _ _ _ _ _ and we need to put 2 heads in those five places, which we can do in 5C2 ways

=> \(\frac{5!}{2!*(5-2)!}\) = \(\frac{5*4*3!}{2!*3!}\) = 10 ways

=> P(2H) = \(\frac{10}{32}\) = \(\frac{5}{16}\)

Quantity B: Probability of getting exactly 3 Heads

We need to get 3 heads out of 5 tosses.

We have 5 places to fill _ _ _ _ _ and we need to put 3 heads in those five places, which we can do in 5C3 ways

=> \(\frac{5!}{3!*(5-3)!}\) = \(\frac{5*4*3!}{3!*2!}\) = 10 ways

=> P(3H) = \(\frac{10}{32}\) = \(\frac{5}{16}\)

Clearly, Quantity A = Quantity B = \(\frac{5}{16}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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