The probability is 1/ that a certain coin will turn up heads on any gi

Question

The probability is 1/2 that a certain coin will turn up heads on any given toss. Anne and Bella have 20 dollars each. Anne tosses the coin five times. Every time the coin turns up heads, Anne gives Bella 2 dollars and every time the coin turns up tails, Bella gives Anne 2 dollars. After five tosses, what is the probability that Bella has more than 20 dollars but less than 30 dollars?

A. 5/32
B. 10/32
C. 15/32
D. 16/32
E. 17/32

A. 5/32
B. 10/32
C. 15/32
D. 16/32
E. 17/32

stne · Answer

Bella meets the criteria ONLY under the following two conditions, in all other cases she falls outside the criteria.

(i) HHHHT - In which case she gets  26  dollars 
(ii) HHHTT - In which case she gets  22  dollars

HHHHT =  (\frac{1}{2})^5*\frac{5!}{4!}  = \frac{5}{32}

HHHTT =  (\frac{1}{2})^5 * \frac{5!}{3! 2!} = \frac{10}{32}

\frac{5}{32}  +  \frac{10}{32}  =  \frac{15}{32}

Ans C

Hope it helped.

GraemeGmatPanda · Answer

Since there are 5 tosses, there are in total 10 dollars exchanged, with 5 scenarios
- Bella loses all 5: she owes 10 dollars --> She finishes with 20 - 10 = 10 dollars (does not meet condition)
- Bella wins 1 and loses 4: She wins 2 dollars but loses 8 --> She finishes with 20 + 2 - 8 = 14 dollars (does not meet condition!)
- Bella wins 2 and loses 3: She wins 4 dollars but loses 8 --> She finishes with 20 + 4 - 6 = 18 dollars (does not meet condition!)
- Bella wins 3 and loses 2: She wins 6 dollars but loses 4 --> She finishes with 20 + 6 - 4 = 22 dollars (meets condition)
- Bella wins 4 and loses 1: She wins 8 dollars but loses 2 --> She finishes with 20 + 8 - 2 = 26 dollars (meets condition)
- Bella wins 5 and loses 0: She wins 10 dollars --> She finishes with 20 + 10 = 30 dollars (does not meet condition)

So we have two scenarios to consider:
- she wins 3 out of 5
- she wins 4 out of 5

3 wins, 2 losses can be written HHHTT but also any other combination of 3 Hs and 2 Ts
P (HHHTT) = \((1/2)^5\) (using the AND rule to multiply each individual probability together)
But we need to then allow for all the 3 Hs and 2 Ts to be rearranged
P (3 wins, 2 losses) = \((1/2)^5\) x \(\frac{5!}{3!2!}\) = \(\frac{10}{32}\)

4 wins, 1 loss can be written HHHHT but also any other combination of 4 Hs and 1 T
P (HHHHT) = \((1/2)^5\)
But we need to then allow for all the 4 Hs to be rearranged
P (4 wins, 1 loss) = \((1/2)^5\) x \(\frac{5!}{4!1!}\) = \(\frac{5}{32}\)

Since we want one condition OR the other, we add the two to get the probability of the answer
Answer = \(\frac{10}{32}\) + \(\frac{5}{32}\) = \(\frac{15}{32}\)

Answer C for Bella to get rich, but not too rich

The probability is 1/ that a certain coin will turn up heads on any gi

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The probability is 1/ that a certain coin will turn up heads on any gi

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