Question:

A man takes a step forward with probability $0.4$ and one step backward with probability $0.6$, then the probability that at the end of eleven steps he is one step away from the starting point is

Updated On: Feb 23, 2024
  • $^{11}C_5\times(0.48)^5$
  • $^{11}C_6\times(0.24)^5$
  • $^{11}C_5\times(0.12)^5$
  • $^{11}C_6\times(0.72)^6$
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The Correct Option is B

Solution and Explanation

A man will be 1 step away from its original position, if he takes 5 steps forward and 6 steps backward or 6 steps forward and 5 steps backward. Hence, required probability $={ }^{11} C_{5}(0.4)^{5}(0.6)^{6}+{ }^{11} C_{6}(0.4)^{6}(0.6)^{5} $ $={ }^{11} C_{5}(0.4)^{5}(0.6)^{5}(0.6+0.4) $ $={ }^{11} C_{5}(0.24)^{5} \cdot 1 $ $={ }^{11} C_{6}(0.24)^{5} \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{n-t}\right]$
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Concepts Used:

Conditional Probability

Conditional Probability is defined as the occurrence of any event which determines the probability of happening of the other events. Let us imagine a situation, a company allows two days’ holidays in a week apart from Sunday. If Saturday is considered as a holiday, then what would be the probability of Tuesday being considered a holiday as well? To find this out, we use the term Conditional Probability.

Let’s discuss certain theorems of Conditional Probability:

  1. Let us consider a random experiment where the sample space S is considered as space and two events namely A and B happen there. Then, the formula would be:

P(S | B) = P(B | B) = 1.

Proof of the same: P(S | B) = P(S ∩ B) ⁄ P(B) = P(B) ⁄ P(B) = 1.

[S ∩ B indicates the outcomes common in S and B equals the outcomes in B].

  1. Now let us consider any two events namely A and B happening in a sample space ‘s’, then, P(A ∩ B) = P(A).

P(B | A), P(A) >0 or, P(A ∩ B) = P(B).P(A | B), P(B) > 0.

This theorem is named as the Multiplication Theorem of Probability.

Proof of the same: As we all know that P(B | A) = P(B ∩ A) / P(A), P(A) ≠ 0.

We can also say that P(B|A) = P(A ∩ B) ⁄ P(A) (as A ∩ B = B ∩ A).

So, P(A ∩ B) = P(A). P(B | A).

Similarly, P(A ∩ B) = P(B). P(A | B).

The interesting information regarding the Multiplication Theorem is that it can further be extended to more than two events and not just limited to the two events. So, one can also use this theorem to find out the conditional probability in terms of A, B, or C.

Read More: Types of Sets

 

Sometimes students get confused between Conditional Probability and Joint Probability. It is essential to know the differences between the two.