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Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, on

Iteru [2.4K]

Answer:

$P(X_i=2) =\dfrac{1}{6}$

$P(X_i=-1) =\dfrac{5}{6}$

Step-by-step explanation:

Given the numbers on the chips = 1, 1, 3 and 5

Miguel chooses two chips.

Condition of winning: Both the chips are same i.e. 1 and 1 are chosen.

Miguel gets $2 on winning and loses $1 on getting different numbers.

To find:

Probability of winning $2 and losing $1 respectively.

Solution:

Here, we are given 4 numbers 1, 1, 3 and 5 out of which 2 numbers are to be chosen.

This is a simple selection problem.

The total number of ways of selecting r numbers from n is given as:

$_nC_r = \frac{n!}{r!(n-r)!}$

Here, n = 4 and r = 2.

So, total number of ways = $_4C_2 = \frac{4!}{2!\times 2!} = 6$

Total number of favorable cases in winning = choosing two 1's from two 1's i.e. $_2C_2 = \frac{2!}{2! 0! } = 1$

Now, let us have a look at the formula of probability of an event E:

$P(E) = \dfrac{\text{Number of favorable ways}}{\text{Total number of ways}}$

So, the probability of winning.

$P(X_i=2) =\dfrac{1}{6}$

Total number of favorable cases for -1: (6-1) = 5

So, probability of getting -1:

$P(X_i=-1) =\dfrac{5}{6}$

Please refer to the attached image for answer table.

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