Probability Test - 60

Given,

A bag contains 5 brown and 4 white socks.

Total number of shocks = 5 + 4 = 9

Now,

P(Both are the same color) = \(\frac{({}^5C_2 + {}^4C_2)}{{}^9C_2}\)

=\(\frac{\frac{5 \times 4}{2 \times 1}+\frac{9 \times 3}{2 \times 1}}{\frac{9 \times 8}{2 \times 1}}\)

= \(\frac{5 \times 4+4 \times 3}{72}\)

= \(\frac{32}{72}\)

= \(\frac{4}{9}\)

Numbers divisible by 4 from 1 to 80 are:

\(4,8,12 \ldots . .80\)

Total no. of numbers divisible by 4 between 1 to 80

\(80=4+(n-1) 4\)

\(80=4 n \)

\(\Rightarrow n=20 \)

\(\therefore\) Required probability \(=\frac{{ }^{20} C_{2}}{{ }^{80} C_{2}}=\frac{19}{316}\)

Let x be number a discrete random variable which denotes the number of heads obtained in n.

n = 8

The general form for the probability of random variable x is,

P(X = x) = ⁿC_x × p^x × q^n-x

Now, in the question, we want at least two heads,

Now, p = q = \(\frac{1}{2}\)

So, \(P(X \geq 2)={ }^{8} C_{2} \times(\frac{1}{2})^{2} \times(\frac{1}{ 2})^{8-2}\)

\(\Rightarrow P(X \geq 2)={ }^{8} C_{2} \times(\frac{1}{2})^{2} \times(\frac{1}{2})^{6}\)

\(\Rightarrow 1-P(X<2)={ }^{8} C_{0} \times(\frac{1}{2})^{0} \times(\frac{1}{2})^{8}+{ }^{8} C_{1} \times(\frac{1}{ 2})^{1} \times(\frac{1}{2})^{8-1}\)

\(\Rightarrow 1-P(X<2)=(\frac{1}{2})^{8}+8 \times(\frac{1}{2})^{1} \times(\frac{1}{ 2})^{7}\)

\(\Rightarrow 1-P(X<2)=\frac{1}{256}+8 \times(\frac{1}{2})^{8}\)

\(\Rightarrow 1-P(X<2)=\frac{1}{256}+\frac{8}{256}\)

\(\Rightarrow 1-P(X<2)=\frac{9 }{256}\)

\(\Rightarrow P(X<2)=1-\frac{9}{256}\)

\(\Rightarrow P(X<2)=\frac{(256-9)}{ 256}\)

\(\Rightarrow P(X<2)=\frac{247 }{256}\)

Given,

A couple has two children.

Let A denotes both children are females i.e. {FF}

Now, P(A) = \(\frac{1}{2}\)×\(\frac{1}{2}\) = \(\frac{1}{4}\)

and B denotes elder children is a female i.e. {FF, FM}

P(B) = \(\frac{1}{4}\) +\(\frac{1}{4}\) = \(\frac{1}{2}\)

Now, P(A ∩ B) = \(\frac{1}{4}\)

Now, P(Both the children are female if elder child is female)

P\((\frac{A}{B})\) = \(\frac{P(A ∩ B)}{P(B)}\)

⇒ P\((\frac{A}{B})\) = \(\frac{(\frac{1}{4})}{(\frac{1}{2})}\)

⇒ P\((\frac{A}{B})\) = \(\frac{1}{2}\)

Given,

A and B are two mutually exclusive events.

So, P(A ∩ B) = 0

Again given P(A) = 0.5 and \( P(\bar B)\) = 0.6

\(P(B)\) = 1 – \( P(\bar B)\) = 1 – 0.6

= 0.4

Now, \(P(A ∪ B)\) = \(P(A)\) + \(P(B)\) – \(P(A ∩ B)\)

⇒ \(P(A ∪ B)\)=\(P(A)\)+\(P(B)\)

⇒ \(P(A ∪ B)\) = 0.5 + 0.4 = 0.9

In a leap year, the total number of days = 366 days.

In 366 days, there are 52 weeks and 2 days.

Now two days may be:

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

Now there is a total of 7 possibilities.

So, total outcomes = 7

In 7 possibilities, Sunday came two times.

So, favorable case = 2

Therefore, the probabilities of getting 53 Sundays in a leap year = \(\frac{2}{7}\)

Let A is the event that 6 does not occur at all.

Now, the probability of at least one 6 occur = 1 – P(A)

= \(1-(\frac{5}{6})^{6}\)

Total cities are 4 i.e. A, B, C, D

Given, Rahul visit four cities, So, n(S) = 4! = 24

Now, sample space is:

S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA, CABD, CADB, CBDA, CDAD, CDAB,CDBA, DABC, DACB, DBCA, DBAC, DCAB, DCBA}

Let G = Rahul visits A first and B last

⇒ G = {ACDB, ADCB}

⇒ n(G) = 2

So, P(G) = \(\frac{n(G)}{n(S)}\) = \(\frac{2}{24}\) = \(\frac{1}{12}\)

Given,

The dice is thrice as likely to show an odd number than an even number. Let the odd number be event \(A\) and the even number be event \(B\).

\(\mathrm{P}(\mathrm{A})=3 \mathrm{P}(\mathrm{B})\)

\(\text {let} P(B)=p \Rightarrow P(A)=3 p\)

Now \(p+3 p=1\)

\(p=\frac{1}{4}\)

When thrown twice, the sum of two numbers thrown is even when both are odd or both are even, i.e.

Possible chances are:

1) Even in 1st throw and Even in 2 nd throw \(=\frac{1}{4} \times \frac{1}{4}\)

\(=\frac{1}{16}\)

2) Odd in 1st throw and Odd in 2nd throw \(=\frac{3}{4} \times \frac{3}{4}\)

\(=\frac{9}{16}\)

Therefore,

The total probability will be:

\(P=\frac{1}{16}+\frac{9}{16}\)

\(P=\frac{1}{16}+\frac{9}{16}\)

\(=\frac{5}{8}\)

Given,

Event \(\mathrm{A}\) is choosing a yellow pencil first, and Event \(\mathrm{B}\) is choosing a yellow pencil second.

Initially, there are 12 pencils, 7 of which are yellow.

Probability the first pencil is yellow \(=\mathrm{P}(\mathrm{A})=\frac{7}{12}\)

If a yellow pencil is chosen, there will be 11 pencils left, 6 of which are yellow.

Probability the second pencil is yellow \(=\mathrm{P}(\mathrm{B})=\frac{6}{11}\)

Given: Two pencils are chosen at random from the box without replacement.

So events are independent of each other.

Probability they are both yellow \(=P(A \cap B)=P(A) \times P(B)\)

\(=\frac{7}{12} \times \frac{6}{11}\)

\(=\frac{7}{22}\)