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Hope it helps.
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Q: what is the probability of getting 3 or 4 heads if a fair coin is flipped 5 times.

total number of possibilities \(= 2 * 2 * 2 * 2 * 2 = 2^5\)

\(P(3) = \frac{5C3}{2^5} = \frac{10}{32}\)

\(P(4) = \frac{5C4}{2^5} = \frac{5}{32}\)

Required probability \(= P(3) + P(4) = \frac{15}{32}\)

Decided to post this question as Manhattan's explanation was unnecessarily cumbersome.

Question asks for the probability of the coin landing tails up either 3 or 4 times = P(3t) + P(4t)

Binomial distribution formula: nCk p^k (1-p)^(n-k)

P(3t) = 5C3 (1/2)^3 (1/2)^2 = 10 (1/2)^5
P(4t) = 5C4 (1/2)^4 (1/2)^1 = 5 (1/2)^5

=> P(3t) + P(4t) = 15/32

y arent we taking into account the probablility of 2 heads or 1

For 2 & 1 heads Kate will end up with < $10 and we want her to win . Therefore, only possibilities are 3 or 4 heads.

I made an educated guess and it worked fine.

Ans- 'D'

MGMAT's anagram helped here as well.

HHHHT = 5!/4!*1! = 5
HHHTT = 5!/3!*2! = 10

Total acceptable cases = 15
Total cases = 32

P = 15/32

total number of possible outcomes when coin will be flipped 5 times= 2^5=32

now, for kate to have more than $10, she must have min. 3 favorable and max. 4 favorable draws in the 5 throws. (because if she wins in all the 5 draws, then she will have $15)

kate will win, every time tails (T) appears on the coin.

case 1) kate wins in 3 draws and losses in two. so we have 3 T's and 2H's(T,T,T,H,H), which can arrange themselves in 5!/(3!)(2!)=10

case 2) kate wins in 4 draws and losses in the 1 draw. so we have 4 T's and 1H (T,T,T,T,H), which can arrange themselves in 5!/4!=5

thus total no. of favorable ways=10+5=15
and thus required probability=15/32

Solved it using Pascal's triangle method and able to got the answer in 30 sec. https://en.wikipedia.org/wiki/Pascal's_triangle

Can anybody explain why the total no.of possibilities is 32 and not 10.

My explanation is
Total outcomes of tossing a coin =2^5 =32

So the probability is exactly half times head and half time tails...i.e.16/32
And since it is asked for 10-15
Answer is 15/32.


Is my approach right??

Posted from my mobile device

In order for Kate to have more than 10 dollars but less than 15 dollars, either of the following two outcomes must have occurred:

T-T-T-T-H, so Kate would have 13 dollars

T-T-T-H-H, so Kate would have 11 dollars

Let’s calculate the probability of each outcome:

P(T-T-T-T-H) = (1/2)^5 = 1/32

Since T-T-T-T-H can be arranged in 5!/4! = 5 ways, the probability is 5/32.

Next:

P(T-T-T-H-H) = (1/2)^5 = 1/32

Since T-T-T-H-H can be arranged in 5!/(3! x 2!) = 10 ways, the probability is 10/32.

So the overall probability that Kate has more than $10 but less than $15 is 15/32.

Answer: D
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cases in which $10<total<$15:
WWWLL=+3-2=+1… total=$11… prob: \(3W2L=(1/2)^5•{arrangements}\frac{5!}{3!2!}=\frac{10}{32}\)
WWWWL=+4-1=+3… total=$13… prob: \(4W1L=(1/2)^5•{arrangements}\frac{5!}{4!1!}=\frac{5}{32}\)
sum cases: \(\frac{10}{32}+\frac{5}{32}=\frac{15}{32}\)

Answer (D)

Why do we need to consider the arrangement here
Please help. its all about getting heads and tails . We need to have 4 heads or 3 heads for kate to have money more than 10 and less than 15
doesnt matter at which turn we are getting head or tail