Probability Test - 59

Given,

Total number of cards = 52

Two cards are lost.

So remaining cards = 50

Now one card is drawn.

The probability that it is a diamond card = \(\frac{13}{50}\)

Now probability that both lost cards are heart = \(\frac{13}{50}\times(\frac{{ }^{11} C _{2}}{{ }^{49} C _{2}})\)

\(=\frac{13}{50}×\frac{(\frac{11×10}{2})}{(\frac{49×48}{2})}\)

\(=\frac{13×11}{{5×49×48}}\)

\(=\frac{143}{{11760}}\)

So the probability that both lost cards are heart is \(\frac{143}{{11760}}\).

The last digit of the four whole numbers can be:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

The chance that any of the four numbers is divisible by 2 or 5 = \(\frac{6}{10}\) = \(\frac{3}{5}\)

Therefore,

The chance that any of the four numbers is not divisible by 2 or 5 = \(1 – \frac{3}{5}\)

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

So, the chance that all of the four numbers are divisible by 2 or 5 = \((\frac{2}{5})×(\frac{2}{5})×(\frac{2}{5})×(\frac{2}{5})\)

= \(\frac{16}{625}\)

This is the chance that the last digit in the product will not be 0, 2, 4, 5, 6, 8 and this is also the chance that the last digit in the product is 1, 3, 7, or 9.

Total number of cases = 6³

= 216

The same number can appear on each of the dice in the following ways:

(1, 1, 1), (2, 2, 2),(3, 3, 3),(4, 4, 4),(5, 5, 5),(6, 6, 6)

So, favorable number of cases = 6

\(\therefore\) Required probability = \(\frac{6}{216}\)

= \(\frac{1}{36}\)

Only two tests are required when the two machines tested are either both working or both are not working.

When both are working:

Probability is = \(\frac{2}{4} \times \frac{1}{3}\)

= \(\frac{1}{6}\)

Both are not working.

Probability is = \(\frac{2}{4} \times \frac{1}{3}\)

= \(\frac{1}{6}\)

Therefore,

Total probability = \(\frac{1}{6}+\frac{1}{6}\)

= \(\frac{1}{3}\)

When two dice are throw, then total outcome = 36

A doubled: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Favourable outcome = 6

Sum is 10: {(4, 6), (5, 5), (6, 4)}

Favourable outcome = 3

Again, A doubled and sum is 10: (5, 5)

Favourable outcome = 1

Now, P(either doubled or a sum of 10 appears) = P(A doubled appear) + P(sum is 10) – P(A doubled appear and sum is 10)

⇒ P(either doubled or a sum of 10 appears) = \(\frac{6}{36} + \frac{3}{36} – \frac{1}{36}\)

= \(\frac{6 + 3 – 1}{36}\)

= \(\frac{8}{36}\)

= \(\frac{2}{9}\)

So, P(neither doubled nor a sum of 10 appears) = 1 – \(\frac{2}{9}\)

= \(\frac{7}{9}\)

When two dice are thrown, then total outcome = 6 × 6 = 36

A: Getting an odd number on the first die.

A = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4),(3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6)}

Total outcome = 18

B: Getting a total of 7 on the two dice.

B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

Total outcome = 6

C: Getting a total of greater than or equal to 8 on the two dice.

C = {(2, 6), (3, 5), (3, 6), (4, 4),(4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Total outcome = 15

Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

n(A ∪ B) = 18 + 6 – 3

\(\Rightarrow\)n(A ∪ B) = 21

\(\therefore\) n(A ∪ B) is 21.

Given,

Number ={1, 2, 3, 4, 5, 6}

Total number of ways of choosing two numbers out of six \(={ }^{6} \mathrm{C}_{2}=\frac{(6 \times 5)}{ 2}\)

\(=3 \times 5=15\)

If smaller number is chosen as 3 then greater has choice are \(4,5,6\)

So, total choices \(=3\)

If smaller number is chosen as 2 then greater has choice are \(3,4,5,6\)

So, total choices \(=4\)

If smaller number is chosen as 1 then greater has choice are \(2,3,4,5,6\)

So, total choices \(=5\)

Total favourable case \(=3+4+5=12\)

Now, required probability \(=\frac{12}{15}\)

\(=\frac{4}{5}\)

Given,

Total number of cards \(=52\)

Number of king card \(=4\)

Now, 7 cards are drawn from 52 cards.

\(P(3 \text { cards are king })=\frac{{ }^{4} C_{3} \times{ }^{48} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{7} }\)

\(=\frac{4 \times\frac{(48 \times 47 \times 46 \times 45)}{(4 \times 3 \times 2 \times 1)\ }}{\frac{(52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46)}{(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} }\)

\(=\frac{4 \times(48 \times 47 \times 46 \times 45) \times(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}{(4 \times 3 \times 2 \times 1) \times(52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46)}\)

\(=\frac{(7 \times 6 \times 5 \times 4 \times 45)}{(52 \times 51 \times 50 \times 49)}\)

\(=\frac{(6 \times 5 \times 4 \times 45)}{(52 \times 51 \times 50 \times 7)}\)

\(=(\frac{(6 \times 4 \times 45)}{(7 \times 52 \times 51 \times 10)}\)

\(=\frac{(6 \times 45)}{(7 \times 13 \times 51 \times 10)}\)

\(=\frac{(6 \times 3)}{(7 \times 13 \times 17 \times 2) }\)

\(=\frac{(3 \times 3) }{(7 \times 13 \times 17) }\)

\(=\frac{9}{1547}\)

Given,

Word is "INSTITUTION"

Total letters \(=11\)

The word contains \(3I,2 N, 1 S, 3 T, 1 U\) and 1O.

Total number of arrangement \(=\frac{11 !}{(3 ! \times 2 ! \times 3 !)}\)

\(=554400\)

Now, taken \(3 \mathrm{~T}\) are together.

So total letter \(=9\)

The number of favorable cases \(=\frac{9 !}{(3 ! \times 2 !)}\)

\(=30240\)

Now, \(P(3\) T are together \()=\frac{30240}{554400}=0.0545\)

One person can select one house out of 3 = ³C₁

= 3

So, three persons can select one house out of three = 3 × 3 × 3 = 27

Thus, the probability that all the three can apply for the same house = \(\frac{3}{27}\)

= \(\frac{1}{9}\)