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Please note that the coin problems are very similar to alphabet problems and can be solved in the same way.

Exactly 4 heads out of 5 would be: HHHHT, HHTHH, or many other combinations.

therefore, no. of possible combinations are 5!/4!*1! = 5

Probability = 5/2^5 = 5/32
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To get 4 heads from 5 tosses there must be 4H and 1T. The probability of getting 4H and then 1T is 1/(2^5).
However, there are 5 ways to arrange 4H and 1T in 5 tosses. This is found using combinations nCr: 5!/(4!)(1!).
Multiplying the probability by the number of arrangements = 5/32.
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Rhoda tosses a coin 5 times. Find the probability of getting exactly 4 heads.

A. 1/32
B. 1/16
C. 3/32
D. 1/8
E. 5/32

The probability of H-H-H-H-T in this order is (½)^4 (½) = 1/32. However, the 4 heads and 1 tail can be arranged in 5!/4! = 5 ways. Therefore, the probability of getting 4 heads and 1 tail is 5 x 1/32 = 5/32.

Answer: E
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Given that Rhoda tosses a coin 5 times and we need to find Find the probability of getting exactly 4 heads.

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

Cases in which we get 4H is same as getting 1T. Cases for 1T can be found by putting 1 tail in any of the 5 slots _ _ _ _ _ => THHHH, HTHHH, HHTHH, HHHTH, HHHHT => 5 ways

=> P(4H) = \(\frac{5}{32}\)

So, Answer will be E
Hope it helps!

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

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