Applied Test Series

Question

A fair coin is tossed 20 times. The probability of getting the three or more heads in a row is 0.7870 and the probability of getting three or more heads in a row or three or more tails in a row is 0.9791. What is the probability of getting three or more heads in a row and three or more tails in a row________

Comments

https://brilliant.org/wiki/probabilistic-principle-of-inclusion-and-exclusion/

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is 0.5949 answer ?

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Kabir5454 Yep, getting the same.

Answer 1

Let probability of getting 3 or more heads in a row = $x$

probability of getting 3 or more tails in a row = $y$

Given $x$ = 0.7870

Since it is a fair coin so, $y$ = $x$ = 0.7870

probability of getting 3 or more heads in a row OR 3 or more tails in a row = P($x$ U $y$) = 0.9791

we have to find probability of getting 3 or more heads AND 3 or more tails in a row = P($x$ ∩ $y$)

P($x$ U $y$) = P($x$) + P($y$) – P($x$ ∩ $y$)

0.9791 = 0.7870 + 0.7870 – P($x$ ∩ $y$)

P($x$ ∩ $y$) = 1.574 – 0.9791 = 0.5949

Answer 2

The probability of getting the three or more heads in a row is 0.7870 ,let it be P(A).

As Probability of Getting head and tail are equally likely, so the probability of getting the three or more Tails in a row is 0.7870. let it be P(B).

Also, the probability of getting three or more heads in a row or three or more tails in a row is 0.9791. This is P(A U B).

The probability of getting three or more heads in a row and three or more tails in a row is P(A) + P(B) -P(A U B)

// From inclusion exclusion principle and addition theorem of Probability.

So,P(A intersection B) = 0.7870 + 0.7870 - 0. 9791 = 0.5949

Answer 3

@Sagar475

Given:

Tosses of the coin = 20 ...

Probability of getting the three or more heads = 0.7870 ...

Probability of getting three or more heads or tails in a row = 0.9791....

To Find:

Probability of getting three or more heads in a row and three or more tails in a row Solution:

Let the event of getting three or more heads in a row in 20 tosses

= X ….

Let the event of getting three or more tails in a row in 20 tosses

= Y ...

Thus, P( X ) = P( Y ) = 0.7870 ...

and

P( X ∪ Y) = 0.9791....

Using the probability relation :

= P(X ∪ Y) = P(X) + P(Y) − P(XY) ….

Thus, the desired probability will be -

P(XY)

= 0.7870 + 0.7870 − 0.9791

= 1.574 - 0.9791

= 0.5949 .…

= 0.6 ( Approx).…

 

The desired probability is  0.6 .…