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A coin is so biased that the probability of falling head when tossed is \[\dfrac{1}{4}\]. If the coin is tossed 5 times the probability of obtaining 2 heads and 3 tails, with heads occurring in succession is:
(a) \[\dfrac{5\times {{3}^{3}}}{{{4}^{5}}}\]
(b) \[\dfrac{{{3}^{3}}}{{{5}^{4}}}\]
(c) \[\dfrac{{{3}^{3}}}{{{4}^{4}}}\]
(d) \[\dfrac{{{3}^{3}}}{{{4}^{5}}}\]

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Last updated date: 20th Jul 2024
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Answer
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Hint: For solving this question you should know about the concept of probability. In this question we will first calculate all the possibilities which are possible according to our question and that will be our complete results and then from that we will select as our requirement. And at the end we will take care of both results. That will be our final answer.

Complete step by step answer:
According to the question it is asked that A coin is so biased that the probability of falling head when tossed is \[\dfrac{1}{4}\]. If the coin is tossed 5 times the probability of obtaining 2 heads and 3 tails is.
So, \[P\left( H \right)=\dfrac{1}{4}\] and \[P\left( T \right)=\dfrac{3}{4}\]
Hence, the required probability will be HHTTT
                   \[=\dfrac{1}{4}.\dfrac{1}{4}.\dfrac{3}{4}.\dfrac{3}{4}.\dfrac{3}{4}\]
Now., consider HHTTT.
Since, HH has to be together, we consider HH as one item.
Hence, the total number of arrangements of 4 items in which 3 are repeated are \[\dfrac{4!}{3!}=4\].
Hence, HHTTT can be internally operated in 4 ways considering that the heads occur in succession. Therefore, the required probability will be \[=\dfrac{{{3}^{3}}}{{{4}^{5}}}.4=\dfrac{{{3}^{3}}}{{{4}^{4}}}\].

So, the correct answer is “Option A”.

Note: For solving these types of questions, always understand the condition clearly. If the conditions are not clear then these will give us the wrong answer. And count every single term according to the given conditions and if there is any term left then the probability will be wrong and that will be wrong to this.