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A certain biased coin when tossed has a probability p of showing head. If the probability that there will be exactly one head in 5 tosses is same as the probability that there will be exactly two heads in 5 tosses, then what is the probability that there will be exactly 3 heads in 5 tosses?
- a)5/16
- b)1/32
- c)40/243
- d)80/243
- e)None of the above
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A certain biased coin when tossed has a probability p of showing head....
The probability that the coin will show tail = 1 - p. Let us first find the probability of getting exactly one head in 5 tosses.
Let only the first toss show head.
Probability o f this event = p x (1 - p)4 But second, third, fourth or fifth tosses can also show heads.
Hence, probability that there is exactly one head in 5 tosses = 5 x p x (1 - p)4
Let only the first toss show head.
Probability o f this event = p x (1 - p)4 But second, third, fourth or fifth tosses can also show heads.
Hence, probability that there is exactly one head in 5 tosses = 5 x p x (1 - p)4
Similarly, let’s find the probability of getting exactly two heads in 5 tosses.
Let only the first and second tosses show head.
Probability = p2 * (1 - p)3 Now, there are 5C2 different ways in which we get exactly 2 heads in 5 tosses.
Let only the first and second tosses show head.
Probability = p2 * (1 - p)3 Now, there are 5C2 different ways in which we get exactly 2 heads in 5 tosses.
Hence, probability that there are exactly 2 heads in 5 tosses = 5C2 x p2 x (1 - p)3 Hence, 5 x p x (1 - p)4 = 5C2 x p2 x (1 -p) 3
1 - p = 2p.
1 - p = 2p.
p = 1/3
Now, let’s find the probability that there are exactly 3 heads in 5 tosses.
Let only first three tosses show head.
Hence, probability = p3 * (1 - p)2 But there are 5C3 different ways in which we can get exactly 3 head in 5 tosses.
Hence, probability of getting exactly 3 heads in 5 tosses = 5C3 x p3 x (1 - p)2
Now, let’s find the probability that there are exactly 3 heads in 5 tosses.
Let only first three tosses show head.
Hence, probability = p3 * (1 - p)2 But there are 5C3 different ways in which we can get exactly 3 head in 5 tosses.
Hence, probability of getting exactly 3 heads in 5 tosses = 5C3 x p3 x (1 - p)2
= 40/243Hence, option 3.
Most Upvoted Answer
A certain biased coin when tossed has a probability p of showing head....
Let's assume that the probability of getting a head on a biased coin is p. Therefore, the probability of getting a tail would be 1-p.
To find the probability of getting exactly one head in 5 tosses, we need to consider the different ways this can happen:
- H T T T T
- T H T T T
- T T H T T
- T T T H T
- T T T T H
The probability of each of these outcomes occurring is p * (1-p) * (1-p) * (1-p) * (1-p), which is equal to p * (1-p)^4.
Similarly, to find the probability of getting exactly two heads in 5 tosses, we need to consider the different ways this can happen:
- H H T T T
- H T H T T
- H T T H T
- H T T T H
- T H H T T
- T H T H T
- T H T T H
- T T H H T
- T T H T H
- T T T H H
The probability of each of these outcomes occurring is p^2 * (1-p) * (1-p) * (1-p), which is equal to p^2 * (1-p)^3.
According to the given information, the probability of getting exactly one head should be equal to the probability of getting exactly two heads:
p * (1-p)^4 = p^2 * (1-p)^3
We can simplify this equation by dividing both sides by p * (1-p)^3:
(1-p) = p
Simplifying further, we get:
1 = 2p
Therefore, p = 1/2.
Now, to find the probability of getting exactly three heads in 5 tosses, we need to consider the different ways this can happen:
- H H H T T
- H H T H T
- H H T T H
- H T H H T
- H T H T H
- H T T H H
- T H H H T
- T H H T H
- T H T H H
- T T H H H
The probability of each of these outcomes occurring is (1/2)^3 * (1/2) * (1/2), which is equal to 1/8 * 1/2 * 1/2, which simplifies to 1/32.
Therefore, the probability of getting exactly three heads in 5 tosses is 1/32, which is option (b).
To find the probability of getting exactly one head in 5 tosses, we need to consider the different ways this can happen:
- H T T T T
- T H T T T
- T T H T T
- T T T H T
- T T T T H
The probability of each of these outcomes occurring is p * (1-p) * (1-p) * (1-p) * (1-p), which is equal to p * (1-p)^4.
Similarly, to find the probability of getting exactly two heads in 5 tosses, we need to consider the different ways this can happen:
- H H T T T
- H T H T T
- H T T H T
- H T T T H
- T H H T T
- T H T H T
- T H T T H
- T T H H T
- T T H T H
- T T T H H
The probability of each of these outcomes occurring is p^2 * (1-p) * (1-p) * (1-p), which is equal to p^2 * (1-p)^3.
According to the given information, the probability of getting exactly one head should be equal to the probability of getting exactly two heads:
p * (1-p)^4 = p^2 * (1-p)^3
We can simplify this equation by dividing both sides by p * (1-p)^3:
(1-p) = p
Simplifying further, we get:
1 = 2p
Therefore, p = 1/2.
Now, to find the probability of getting exactly three heads in 5 tosses, we need to consider the different ways this can happen:
- H H H T T
- H H T H T
- H H T T H
- H T H H T
- H T H T H
- H T T H H
- T H H H T
- T H H T H
- T H T H H
- T T H H H
The probability of each of these outcomes occurring is (1/2)^3 * (1/2) * (1/2), which is equal to 1/8 * 1/2 * 1/2, which simplifies to 1/32.
Therefore, the probability of getting exactly three heads in 5 tosses is 1/32, which is option (b).
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A certain biased coin when tossed has a probability p of showing head. If the probability that there will be exactly one head in 5 tosses is same as the probability that there will be exactly two heads in 5 tosses, then what is the probability that there will be exactly 3 heads in 5 tosses?a)5/16b)1/32c)40/243d)80/243e)None of the aboveCorrect answer is option 'C'. Can you explain this answer?
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