Is This the Biased Coin? Part 2

Part 1

There are 2 coins on a table. One of them lands on heads with probability 50%, and the other coin lands on heads with probability 75%, but you don't know which is which. You pick up one coin at random and flip it. It lands on heads. If you flip the same coin again, what is the probability that it will land on heads again?

Write your answer as a percentage.


The answer is 65.000.

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3 solutions

Jesse Li
Nov 27, 2020

Let's work from the beginning to the end and consider which coin you chose, the result of the first flip, and the result of the second flip. We are only concerned with the cases where the result of the first flip was heads. Therefore, there are four possible cases:

  1. You picked the 75% coin and flipped two heads (Probability: 0.5 × 0.75 × 0.75 = 0.28125 0.5 \times 0.75\times 0.75=0.28125 )
  2. You picked the 50% coin and flipped two heads (Probability: 0.5 × 0.5 × 0.5 = 0.125 0.5\times 0.5\times 0.5=0.125 )
  3. You picked the 75% coin and flipped heads and then tails (Probability: 0.5 × 0.75 × 0.25 = 0.09375 0.5\times 0.75\times 0.25=0.09375 )
  4. You picked the 50% coin and flipped heads and then tails (Probability: 0.5 × 0.5 × 0.5 = 0.125 0.5\times 0.5\times 0.5=0.125 )

The first two cases fulfill the condition that the second flip results in heads. So, to find the probability that the second flip results in heads given that the first flip was heads, we take the sum of the probabilities of the first two cases and divide by the total of these four probabilities. So, the answer is 0.28125 + 0.125 0.28125 + 0.125 + 0.09375 + 0.125 = 0.65 = 65 % \frac{0.28125+0.125}{0.28125+0.125+0.09375+0.125}=0.65=\boxed{65\%}

Why are they being multiplied by 0.5 tho? Isn’t their probability already given? (75 and 50)

Hasnain Ashraf - 6 months, 1 week ago

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Because you have a 50% chance of picking each coin

Jesse Li - 6 months, 1 week ago
Gediminas Sadzius
Nov 28, 2020

If you pick one coin at random and flip it, and it lands on heads (H), the probability that you have picked the fair (F) coin is:

P ( F H ) = P(F|H)= P ( F ) × P ( H F ) P ( H ) \frac{P(F)×P(H|F)}{P(H)}

where,

P(F) - probability of picking the fair coin, = 1 2 \frac{1}{2}

P(H|F) - probability of flipping heads once you've picked the fair coin, = 1 2 \frac{1}{2}

P(H) - the total probability of flipping heads:

P ( H ) = P(H)= P ( F ) × P ( H F ) + P ( B ) × P ( H B ) P(F)×P(H|F)+P(B)×P(H|B)

where,

P(B) - probability of picking the biased (B) coin, = 1 2 \frac{1}{2}

P(H|B) - probability of flipping heads once you've picked the biased coin, = 3 4 \frac{3}{4}

So if you flip the same coin again, the probability that it lands on heads is:

P = P= P ( F H ) × P ( H F ) + ( 1 P ( F H ) ) × P ( H B ) P(F|H)×P(H|F)+(1-P(F|H))×P(H|B)

Putting the numbers together, the probability that the coin lands on heads again is (2/5 x 1/2 + (1 - 2/5) x 3/4) x 100 = 65 % \boxed{65\%}

Richard Desper
Nov 30, 2020

Let C 1 C_1 be the event "used first coin" and C 2 C_2 be the event "used second coin". And let H 1 H_1 be the event "tossed heads on first toss" and H 2 H_2 be the event "tossed heads on second toss".

We are interested in P [ H 2 H 1 ] P[H_2|H_1] .

But since the coin toss probabilities depend on which coin was used, we need to express P [ H 2 H 1 ] P[H_2|H_1] as the sum P [ H 2 C 1 ] P [ C 1 H 1 ] + P [ H 2 C 2 ] P [ C 2 H 1 ] P[H_2|C_1]*P[C_1|H_1] + P[H_2|C_2]*P[C_2|H_1] Thus we need to know what P [ C 1 H 1 ] P[C_1|H_1] and P [ C 2 H 1 ] P[C_2|H_1] are.

First note that P [ H 1 ] = P [ C 1 ] P [ H 1 C 1 ] + P [ C 2 ] P [ H 1 C 2 ] = 0.5 0.5 + 0.5 0.75 = 0.625 P[H_1] = P[C_1]*P[H_1|C_1] + P[C_2]*P[H_1|C_2] = 0.5*0.5 + 0.5*0.75 = 0.625 , and that P [ C 1 H 1 ] = 0.25 P[C_1H_1] = 0.25 while P [ C 2 H 1 ] = 0.375 P[C_2H_1] = 0.375 .

Now P [ C 1 H 1 ] = P [ C 1 H 1 ] P [ H 1 ] = 0.25 0.625 = 0.4 P[C_1|H_1] = \frac{P[C_1 H_1]}{P[H_1]} = \frac{0.25}{0.625} = 0.4 and P [ C 2 H 1 ] = P [ C 2 H 1 ] P [ H 1 ] = 0.375 0.625 = 0.6 P[C_2|H_1] = \frac{P[C_2 H_1]}{P[H_1]} = \frac{0.375}{0.625} = 0.6 .

With these probabilities in hand, we can calculate

P [ H 2 H 1 ] = P [ H 2 C 1 ] P [ C 1 H 1 ] + P [ H 2 C 2 ] P [ C 2 H 1 ] P[H_2|H_1] = P[H_2|C_1]*P[C_1|H_1] + P[H_2|C_2]*P[C_2|H_1]

= 0.5 0.4 + 0.75 0.6 = 0.2 + 0.45 = 0.65. = 0.5*0.4 + 0.75*0.6 = 0.2 + 0.45 = 0.65.

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