Question

Number 1 to 10 are written on ten separate slips ( one number on one slip) kept in a box and mixed well. One slip is chosen from The box without looking in to it. What is The probability of getting a number 6 and a 4 ?
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Answer 1

Answer:

Probability of getting 6 is 1/10

and probability of getting 4 is 1/10

Answer 2

Since the numbers from 1 to 10 are written in 10 slips, each number on each slip, and only one slip is chosen from them at a time, the possible sample space is,

$\longrightarrow\sf{S=\{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10\}}$

Therefore,

$\longrightarrow\sf{n(S)=10}$

Let,

• $\sf{A=}$ event of getting a number 6.

• $\sf{B=}$ event of getting a number 4.

Then we see that A and B are simple events.

$\longrightarrow\sf{A=\{6\}}$

$\longrightarrow\sf{B=\{4\}}$

So that,

$\longrightarrow\sf{n(A)=n(B)=1}$

Then the probability of getting a number 6 is,

$\longrightarrow\sf{P(A)=\dfrac{1}{10}}$

And that of getting a number 4 is,

$\longrightarrow\sf{P(B)=\dfrac{1}{10}}$

The event of getting a number 6 or 4 is,

$\longrightarrow\sf{A\cup B=\{4,\ 6\}}$

So,

$\longrightarrow\sf{n(A\cup B)=2.}$

Then the probability of getting a number 6 or 4 is,

$\longrightarrow\sf{P(A\cup B)=\dfrac{2}{10}}$

Hence the probability of getting a number 4 and 6 is,

$\longrightarrow\sf{P(A\cap B)=P(A)+P(B)-P(A\cup B)}$

$\longrightarrow\sf{P(A\cap B)=\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{2}{10}}$

$\longrightarrow\large\text{ \sf{\underline{\underline{P(A\cap B)=0}}} }$

Since one slip is chosen at a time and not two or more, there's no probability of getting 6 and 4 simultaneously. Hence the answer is 0.

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If two slips were taken at a time then the total no. of outcomes would be,

$\longrightarrow\sf{n(S)=\ ^{10}C_2=45}$

and the no. of events of getting 6 and 4 would be,

$\longrightarrow\sf{n(A\cap B)=\ ^1C_1\times\ ^1C_1=1}$

Hence the probability would be,

$\longrightarrow\sf{P(A\cap B)=\dfrac{1}{45}}$