Probability Test - 61

Given,

A bag contains 10 black and 20 white balls

So, Total number of outcomes \(=10+20\)

\(=30\) balls

Now, one ball is drawn at random and it will be white

So, Number of ways \(={ }^{20} \mathrm{C}_{1}\)

\(=20\)

Probability that ball is white \(=\frac{20}{30}\)

\(=\frac{2}{3}\)

Given,

Number of red balls = 6

Number of white balls = 5

Number of black balls = 4

Total number of balls \(=5+4+6\)

\(=15\)

Selection of 1 ball out of \(15={ }^{15} \mathrm{C}_{1}\)

Selection of 2nd ball out of\(14={ }^{14} \mathrm{C}_{1}\)

Selection of 1 White ball out of\(5={ }^{5} \mathrm{C}_{1}\)

Selection of 2nd White ball out of\(4={ }^{4} \mathrm{C}_{1}\)

So,

The required probability \((P)=\frac{{ }^{5} \mathrm{C}_{1}×{ }^{4} \mathrm{C}_{1}}{{}^{15} \mathrm{C}_{1}×{ }^{14} \mathrm{C}_{1}}\)

\(P=\frac{5 \times 4}{15 \times 14}\)

\(=\frac{2}{21}\)

There are 26 red cards out of total of 52 cards which also include 2 kings

So the probability of getting a red card \(\left(P_{1}\right)=\frac{26}{52}\)

Now from 4 kings as 2 kings are already counted there 2 kings are left

So the probability of getting either of them \(\left(\mathrm{P}_{2}\right)=\frac{2}{52}\)

Therefore,

The probability that the card is red colour or king \((P)=P_{1}+P_{2}\)

\(P=\frac{26+2}{52}\)

\(P=\frac{28}{52}=\frac{7}{13}\)

Let's say that \(\mathrm{A}\) is the event of drawing a 10 and \(\mathrm{B}\) be the event of drawing a spade.

Probability of drawing a \(10=\mathrm{P}(\mathrm{A})=\frac{4}{52}\).

Probability of drawing a spade \(=\mathrm{P}(\mathrm{B})=\frac{13}{52}\)

Probability of drawing a 10 of spade \(=\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{52}\)

Therefore,

Probability of drawing a 10 or a spade \(P(A \cup B)=P(A)+P(B)-P(A \cap B)\)

\(=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}\)

\(=\frac{16}{52}=\frac{4}{13}\)

Given,

A bag contains 4 white and 2 black balls and another bag contains 3 white and 5 black balls.

We have to find probability that one ball is white and one ball is black.

Possible cases:

Let \(\mathrm{E}_{1}=\) Black ball from bag 1 and white ball from bag 2 and \(\mathrm{E}_{2}=\) White ball from bag 1 and black ball from bag 2 .

\(P\) (one ball is white and one ball is black) \(=P\left(E_{1}\right)+P\left(E_{2}\right)\)

\(=\frac{2}{6} \times \frac{3}{8}+\frac{4}{6} \times \frac{5}{8}\)

\(=\frac{26}{48}=\frac{13}{24}\)

Given,

There are 5 white (W) socks, 4 red (R) socks and 3 green (G) socks

Total socks \(=5+4+3\)

\(=12\)

Probability of \(1^{\text {st }}\) sock white \(=\frac{5}{12}\)

Probability of \(2^{\text {nd }}\) sock white \(=\frac{4}{11}\)

Probability of two white socks \(=(\frac{5}{12}) \times(\frac{4}{11})\)

\(=\frac{5}{33}\)

Similarly, Probability of two red \(=(\frac{4}{12}) \times(\frac{3}{11})\)

\(=\frac{1}{11}\)

Probability of two green \(=(\frac{3}{12}) \times(\frac{2}{11})\)

\(=\frac{1}{22}\)

We want socks of same color i.e., 2 white or 2 red or 2 green

Total probability \(=\frac{5}{33}+\frac{1}{11}+\frac{1}{22}\)

\(=\frac{1}{11}\left(\frac{5}{3}+1+\frac{1}{2}\right)\)

\(=\frac{1}{11}\left(\frac{19}{6}\right)\)

\(=\frac{19}{66}\)

Given,

We are picking the four balls without replacement and that the 4 white and 4 red balls are the only balls in the box and that we choose 2 red balls and 2 white balls.

Number of ways to choose 4 balls out of \(8={ }^{8} \mathrm{C}_{4}\)

\(=\frac{8 !}{4 ! 4 !}=70\)

Number of ways to pick 2 red balls out of \(4={ }^{4} \mathrm{C}_{2}\)

\(=\frac{4 !}{2 ! 2 !}=6\)

Number of ways to pick 2 white balls out of \(4={ }^{4} \mathrm{C}_{2}\)

\(=\frac{4 !}{2 ! 2 !}=6\)

Probability \(=\frac{6 \times 6}{70}=\frac{36}{70}\)

\(=\frac{18}{35}\)

Probability that \(\mathrm{A}\) wins \(=\frac{6}{12}=\frac{1}{ 2}\) 

Probability that \(\mathrm{B}\) wins \(=\frac{4}{12}=\frac{1}{3}\)

Now, to win in the series of 3 games, one needs to win two times

If \(\mathrm{A}\) win, probability \(=\mathrm{W}_{\mathrm{A}} \times \mathrm{W}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}=\frac{1}{2} \times \frac{1}{3} \times \frac{1}{2}=\frac{1}{12}\) and

If B win, probability \(=W_{B} \times W_{A} \times W_{B}\)

\(=\frac{1}{3 }\times \frac{1}{ 2} \times \frac{1 }{3}=\frac{1}{18}\)

So, required probability \(=\frac{1 }{12}+\frac{1}{18}\)

\(=\frac{5}{36}\)

The total number of caps \(=2+4+5+1=12\)

The number of ways for selecting four caps \(={ }^{12} \mathrm{C}_{4}\)

The total number of caps other than green \(=12-5=7\)

The number of ways for selecting four caps other than green \(={ }^{7} \mathrm{C}_{4}\)

The probability of selecting four caps and none of them are green \(\mathrm{P}\left(\mathrm{A}^{\prime}\right)=\) \(\frac{\text{The number of ways for selecting four caps other than green}}{\text{The number of ways for selecting four caps}}\)

\(\Rightarrow\mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{{ }^{7} \mathrm{C}_{4}}{{ }^{12} \mathrm{C}_{4}}\)

\(\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{7 !(12-4) ! 4 !}{4 !(7-4) ! 12 !}\)

\(\Rightarrow \mathbf{P}\left(\mathbf{A}^{\prime}\right)=\frac{7}{99}\)

Given,

Measures of rectangule are = 6 inches and 5 inches

Total area \(=6 \times 5=30\) sq. inch

Now, the randomly selected point is at least one inch from the edge of the rectangle

So, by leaving space of 1 inch inside the original rectangle will give another rectangle of length 4 inch and breadth 3 inch

So, area of required rectangle \(=4 \times 3=12 \mathrm{sq}\). inch

Probability \(=\frac{\text { required area }}{\text { total area }}\)

\(=\frac{12}{30}\)

\(=\frac{2}{5}\)