Statement I A fair coin is tossed four times and a person win â¹ 1 for each head and lose â¹ 1.50 for each tail that turns up. From the sample space, how many different amounts of money you can have after four tosses and then the probability of having each of these amounts is (1/16), (1/4), (3/8), (1/4), (1/16) respectively? Statement II If (2/11) is the probability of an event, then the probability of the event 'not A ' is (9/11).

Question

Statement I A fair coin is tossed four times and a person win â¹ 1 for each head and lose â¹ 1.50 for each tail that turns up. From the sample space, how many different amounts of money you can have after four tosses and then the probability of having each of these amounts is (1/16), (1/4), (3/8), (1/4), (1/16) respectively? Statement II If (2/11) is the probability of an event, then the probability of the event 'not A ' is (9/11). (A) Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I (B) Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I (C) Statement I is true; Statement II is false (D) Statement I is false; Statement II is true

Tardigrade Admin · Accepted Answer

I. If a coin is tossed four times, then total number of possible outcomes

= 2^{4} = 16

For these cases, sample space can be written as

Sample space	Amount
HHHH	$1 + 1 + 1 + 1 = 4$
HH HT	$1 + 1 + 1 + - 1.50 = 3 - 1.50 = 1.50$
HH TH	$1 + 1 - 1.50 + 1 = 3 - 1.50 = 1.50$
HHTT	$1 + 1 - 1.50 - 1.50 = 2 - 3 = - 1.00$
HTHH	$1 - 1.50 + 1 + 1 = 3 - 1.50 = 1.50$
HTHT	$1 - 1.50 + 1 - 1.50 = 2 - 3 = - 1.00$
HTTH	$1 - 1.50 - 1.50 + 1 = 2 - 3 = - 1.00$
HTTT	$1 - 1.50 - 1.50 - 1.50 = 1 - 4.50 = - 3.50$
THHH	$- 1.50 + 1 + 1 + 1 = - 1.50 + 3 = 1.50$
THHT	$- 1.501 + 1 - 1.50 = 2 - 3.00 = - 1.00$
THTH	$- 1.50 + 1 - 1.50 + 1 = 2 - 3.00 = - 1.00$
THTT	$- 1.50 + 1 - 1.50 - 1.50 = 1 - 4.50 = - 3.50$
TTHH	$- 1.50 - 1.50 + 1 + 1 = 2 - 3.00 = - 1.00$
TTHT	$- 1.50 - 1.50 + 1 - 1.50 = 1 - 4.50 = - 3.50$
TTTH	$- 1.50 - 1.50 - 1.50 + 1 = - 4.50 + 1 = - 3.50$
TTTT	$- 1.50 - 1.50 - 1.50 - 1.50 = - 6.00$

Hence, from above sample space, we get five types of different amounts i.e.,

4, 1.50, - 1.00, - 3.50, - 6.00

i.e.,

Amounts	Number of times occurance
4.00	1
1.50	4
-1.00	6
-3.50	4
-6.00	1
Total	16

(negative sign indicates the losing value of amount)

\Rightarrow P (winning ₹ 4.00)

= \frac{Number of favourable outcomes}{Total number of outcomes} = \frac{1}{16}

\Rightarrow P (having ₹1.50) = \frac{4}{16} = \frac{1}{4}

\Rightarrow P (having ₹ - 1.00) = \frac{6}{16} = \frac{3}{8}

\Rightarrow P (having ₹ - 3.50) = \frac{4}{16} = \frac{1}{4}

\Rightarrow P (having ₹ - 6.00) = \frac{1}{16}

II. If

A

be any event, then given

P (A) = \frac{2}{11}

∴ P (not A) = P (A^{'}) = 1 - P (A)

= 1 - \frac{2}{11} = \frac{9}{11}

[∵

sum of probabilities of an event

(A)

and its complementary event

(A^{'})

is always 1]